MATH 110 - INTERMEDIATE
ALGEBRA - Applications of quadratic
equations
Question Bank - Set 3
Liberty University
Question 1
Question
A rectangular garden has an area of 120 square meters. The length is 5 meters
longer than the width. Find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Since the length is 5 meters longer
than the width, the length is x+ 5 meters.
Step 1: Write the equation for the area of the garden. The area of a rectangle
is given by the formula A= length ×width. Therefore, we have:
x(x+ 5) = 120
Step 2: Expand and rearrange the equation.
x2+ 5x= 120
Step 3: Rewrite the equation in standard form.
x2+ 5x−120 = 0
Step 4: Factor the quadratic equation.
(x+ 15)(x−8) = 0
Step 5: Find the values of x. Setting each factor to zero gives us:
x+ 15 = 0 or x−8=0
x=−15 or x= 8
Step 6: Discard the negative value since we are dealing with lengths. So,
the width of the garden is 8 meters and the length is 8 + 5 = 13 meters.
Therefore, the dimensions of the garden are 8 meters by 13 meters.
Question 2
Question
A rectangular garden has a fixed perimeter of 60 meters. The length of the
garden is 4 meters longer than its width. Find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Then, the length of the garden is x+4
meters.
Step 1: Write an equation based on the information given. The perimeter
of a rectangle is given by 2 ×(length + width). Since the perimeter is 60 meters,
we have:
2×(x+ (x+ 4)) = 60
Step 2: Solve the equation for x.
2×(2x+ 4) = 60
4x+ 8 = 60
4x= 52
x= 13
Step 3: Find the dimensions of the garden. The width, x, is 13 meters.
The length, x+ 4, is 13 + 4 = 17 meters.
Step 4: Check the answer. The perimeter of the garden is 2 ×(13 + 17) =
2×30 = 60 meters, which matches the given fixed perimeter. Therefore, the
dimensions of the garden are width: 13 meters and length: 17 meters.
Question 3
Question
A rectangular garden has a fixed perimeter of 40 meters. If the area of the
garden is to be maximized, what should the dimensions of the garden be?
Solution
Step 1: Let xand yrepresent the dimensions of the garden. Step 2: Write the
equation for the perimeter of the garden in terms of xand y. Step 3: Write
the equation for the area of the garden in terms of xand y. Step 4: Express
yin terms of xin the equation for the area. Step 5: Find the critical points
by taking the derivative of the area equation with respect to xand setting it
equal to 0. Step 6: Determine if it is a maximum or minimum point by using
the second derivative test. Step 7: Calculate the dimensions of the garden at
the maximum area.
2
Let’s find the dimensions of the garden that maximize the area.
Step 1: Let xand yrepresent the dimensions of the garden.
Step 2: The perimeter of a rectangle is given by 2x+2y= 40, or simplifying,
x+y= 20.
Step 3: The area of a rectangle is given by A=xy.
Step 4: Since y= 20 −x, substitute this into the area equation to get
A(x) = x(20 −x) = 20x−x2.
Step 5: Find the critical points by taking the derivative of A(x) and setting
it equal to 0:
dA
dx = 20 −2x= 0
20 = 2x
x= 10
Step 6: Use the second derivative test to check if x= 10 is a maximum:
d2A
dx2=−2<0
Since the second derivative is negative, the point is a maximum.
Step 7: Now that we have x= 10, substitute it back into y= 20 −x:
y= 20 −10
y= 10
Therefore, the dimensions of the garden with the maximum area are 10
meters by 10 meters.
Question 4
Question
A rectangular garden is to be created using a wall as one of its sides. Three
sides of the garden are to be enclosed with a fence, while the fourth side is the
existing wall. If the area of the garden is 200 square meters and the length of
the garden is 10 meters longer than its width, find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Then the length of the garden is x+10
meters.
The area of a rectangle is given by the formula A= width ×length. We are
given that the area of the garden is 200 square meters, so we have the equation:
x(x+ 10) = 200
3
Expanding and simplifying this equation, we get:
x2+ 10x−200 = 0
To solve this quadratic equation, we can either factorize it or use the quadratic
formula. Let’s use the quadratic formula: The quadratic formula states that for
an equation of the form ax2+bx +c= 0, the solutions are given by:
x=−b±√b2−4ac
2a
In this case, a= 1, b= 10, and c=−200. Plugging these values into the
formula, we get:
x=−10 ±p102−4(1)(−200)
2(1)
x=−10 ±√100 + 800
2
x=−10 ±√900
2
Therefore, we have two possible solutions for the width of the garden:
x=−10 + 30
2=20
2= 10
x=−10 −30
2=−40
2=−20
Since the width cannot be negative, we take x= 10.
Thus, the width of the garden is 10 meters and the length is 10 + 10 = 20
meters. So, the dimensions of the garden are 10 meters by 20 meters.
Question 5
Question
A rectangular garden has a length that is 6 meters longer than its width. If the
area of the garden is 300 square meters, find the dimensions of the garden.
Solution
1. Let’s denote the width of the garden as wmeters. We know that the length of
the garden is 6 meters longer, so the length can be represented as w+ 6 meters.
2. The area of a rectangle is given by the formula A= length ×width.
Substituting our expressions for length and width into this formula, we have:
w(w+ 6) = 300
4
3. Now, we can simplify the equation by expanding the product on the left
side:
w2+ 6w= 300
4. Rearranging the equation to set it equal to zero, we get a quadratic
equation:
w2+ 6w−300 = 0
5. To solve this quadratic equation, we can factor it. We look for two
numbers that multiply to -300 and add to 6. These numbers are 20 and -15.
So, we can rewrite the equation as:
w2+ 20w−15w−300 = 0
6. Factor by grouping:
w(w+ 20) −15(w+ 20) = 0
(w−15)(w+ 20) = 0
7. By setting each factor to zero, we find the possible values for the width:
w−15 = 0 or w+ 20 = 0
w= 15 or w=−20
8. Since the width cannot be negative, we discard w=−20 and conclude
that the width is w= 15 meters.
9. Using this width, we can find the length:
length = w+ 6 = 15 + 6 = 21
10. Therefore, the dimensions of the garden are a width of 15 meters and a
length of 21 meters.
Question 6
Question
A rectangular garden has a perimeter of 100 meters. The length of the garden
is 10 meters longer than its width. Find the dimensions of the garden.
Solution
Let’s denote the width of the garden as xmeters. Since the length is 10 meters
longer than the width, the length can be represented as x+ 10 meters.
Step 1: Write down the equation based on the given information. The
perimeter of a rectangle is given by 2×(length + width). From the problem, we
5
know that the perimeter is 100 meters. Therefore, the equation can be written
as:
2(x+x+ 10) = 100
Step 2: Simplify the equation.
4x+ 20 = 100
4x= 80
x= 20
Step 3: Find the length. The length of the garden is x+ 10 meters.
x+ 10 = 20 + 10 = 30
Step 4: State the dimensions of the garden. Therefore, the width of the
garden is 20 meters and the length is 30 meters. The dimensions of the garden
are 20 meters by 30 meters.
Question 7
Question
A rectangular flower garden is 12 meters long and 8 meters wide. The garden is
to be enlarged by increasing both the length and the width by the same amount.
If the area of the new garden is 208 square meters more than the area of the
original garden, find the dimensions of the new garden.
Solution
Let xbe the amount by which both the length and the width are increased.
Step 1: Determine the area of the original garden. The area of the original
garden is given by:
Areaoriginal = length ×width = 12 ×8 = 96 sq meters
Step 2: Determine the area of the new garden. After increasing both the
length and the width by x, the new dimensions are 12 + xmeters and 8 + x
meters. The area of the new garden is then:
Areanew = (12 + x)(8 + x) = 96 + 20x+x2
Step 3: Set up an equation based on the given information. Since the new
garden is 208 square meters larger than the original garden, we have:
Areanew = Areaoriginal + 208
96 + 20x+x2= 96 + 208
6
Step 4: Solve the equation.
x2+ 20x−208 = 0
Factoring the quadratic equation, we get:
(x+ 26)(x−8) = 0
So, x=−26 or x= 8. Since the dimensions cannot be negative, the amount by
which the garden is enlarged is 8 meters.
Step 5: Determine the dimensions of the new garden. The dimensions of
the new garden are: Length: 12 + 8 = 20 meters Width: 8 + 8 = 16 meters
Therefore, the new garden is 20 meters long and 16 meters wide.
Question 8
Question
A company manufactures and sells a certain product. The revenue from selling
xunits of the product is given by the function R(x) = −2x2+100x. The cost of
producing xunits of the product is given by the function C(x)=0.5x2+40x+10.
Find:
1. The profit function P(x).
2. The number of units that must be sold to maximize profit.
3. The maximum profit possible.
Solution
1. To find the profit function P(x), we subtract the cost function from the
revenue function:
P(x) = R(x)−C(x)=(−2x2+ 100x)−(0.5x2+ 40x+ 10)
Simplify to get:
P(x) = −2x2+ 100x−0.5x2−40x−10 = −2.5x2+ 60x−10
Therefore, the profit function is P(x) = −2.5x2+ 60x−10.
2. To find the number of units that must be sold to maximize profit, we first
need to find the critical points of the profit function P(x). The profit function is
a quadratic function, and its maximum will occur at the vertex of the parabola.
The x-coordinate of the vertex of a parabola in the form ax2+bx +cis given
by −b
2a. In this case, a=−2.5 and b= 60. Therefore, the number of units to
be sold to maximize profit is:
x=−60
2(−2.5) =−60
−5= 12
7
.
3. To find the maximum profit possible, substitute x= 12 into the profit
function:
P(12) = −2.5(12)2+ 60(12) −10 = −360 + 720 −10 = 350
Therefore, the maximum profit possible is 350.
Question 9
Question
A photographer is hired to take wedding photos at a rate of xdollars per
photo. The photographer takes a total of 50 photos during the wedding. If
the photographer is paid a total of $500 for the photos, find the rate per photo.
Solution
Let xbe the rate per photo in dollars.
Step 1: Write an equation using the given information. The total pay for 50
photos is given by the product of the rate per photo and the number of photos
taken. Since the total pay is $500, we have:
50x= 500
Step 2: Solve the equation to find the rate per photo. Divide both sides of
the equation by 50:
x=500
50
x= 10
Step 3: Answer The rate per photo is $10.
Question 10
Question
A rectangular garden is to be created using 60 meters of fencing. If one side of
the garden is against a wall, find the dimensions of the garden that maximize
the area.
Solution
Let xbe the length of the side of the garden against the wall and ybe the width
of the garden.
Step 1: Express the total perimeter in terms of xand y. The perimeter of
the garden is given as 60 meters, so we have the equation: 2x+y= 60.
8
Step 2: Express the area of the garden in terms of xand y. The area of
the garden is given by A=xy.
Step 3: Express yin terms of x. From the perimeter equation, we can
express yas y= 60 −2x.
Step 4: Substitute y= 60−2xinto the area formula. Substitute y= 60−2x
into the area formula A=xy to get A=x(60 −2x), which simplifies to A=
60x−2x2.
Step 5: Find the critical points by taking the derivative of the area function.
To find the dimensions that maximize the area, we need to find the critical points
by taking the derivative of the area function: dA
dx = 60 −4x.
Step 6: Set the derivative equal to zero and solve for x. Set dA
dx = 60−4x= 0
and solve for x: 60 = 4x=⇒x= 15.
Step 7: Find the corresponding value of y. To find the corresponding value
of y, substitute x= 15 into y= 60 −2x:y= 60 −2(15) = 60 −30 = 30.
Step 8: Check the endpoints. Check the endpoints of the possible interval
to ensure the maximum is at x= 15.
Step 9: State the dimensions that maximize the area. Therefore, the di-
mensions that maximize the area are x= 15 meters (length) and y= 30 meters
(width).
Question 11
Question
A rectangular playground has a length that is 10 meters longer than its width.
If the area of the playground is 192 square meters, find the dimensions of the
playground.
Solution
Let’s denote the width of the playground as xmeters. Since the length is 10
meters longer than the width, we can express the length as x+ 10 meters.
We know that the area of a rectangle is given by the formula A= length ×
width. For this playground, the area is given as 192 square meters:
x(x+ 10) = 192
Expanding the left side of the equation:
x2+ 10x= 192
Rearranging the equation into standard form:
x2+ 10x−192 = 0
Next, we can solve this quadratic equation using the quadratic formula:
9
x=−b±√b2−4ac
2a
where a= 1, b= 10, and c=−192.
Plugging these values into the quadratic formula:
x=−10 ±√102−4×1× −192
2×1
x=−10 ±√100 + 768
2
x=−10 ±√868
2
x=−10 ±2√217
2
x=−5±√217
Therefore, the possible values for the width are −5 + √217 and −5−√217.
Since the width cannot be negative, the correct value for the width is −5+√217.
Now, we can find the length by substituting the width back into the expres-
sion x+ 10:
Length = (−5 + √217) + 10
Length = 5 + √217
So, the width of the playground is approximately 1.78 meters and the length
is approximately 15.78 meters.
Question 12
Question
A rectangular garden is four times as long as it is wide. If the area of the garden
is 100 square meters, find the dimensions of the garden.
Solution
Let’s denote the width of the garden as wmeters. Since the length of the garden
is four times the width, the length can be represented as 4wmeters. We are
given that the area of the garden is 100 square meters.
Step 1: Write the equation for the area of the garden in terms of w. The
area of a rectangle is given by the formula A= length ×width. Substituting
the given values, we have:
100 = 4w×w
10
Step 2: Simplify the equation to form a quadratic equation.
100 = 4w2
Step 3: Rewrite the equation in standard quadratic form.
4w2−100 = 0
Step 4: Solve the quadratic equation by factoring or using the quadratic
formula. We can factor out a 4 from the equation:
4(w2−25) = 0
Step 5: Solve for wby setting each factor to zero. We have two possible
solutions:
w2−25 = 0
Step 6: Write down the two possible solutions for w.
w2−25 = 0
w2= 25
w=±5
Since the width cannot be negative, the width of the garden is 5 meters.
Step 7: Find the length of the garden. The length of the garden is 4 times
the width, which is:
4×5 = 20
Step 8: Write the final answer. The dimensions of the garden are width =
5 meters and length = 20 meters.
Question 13
Question
A cargo plane flying at a constant speed releases a box of supplies at a height of
500 meters above the ground. The horizontal distance between the plane and
the point directly below the box when the box is released is 800 meters. Find
the horizontal distance the box travels before hitting the ground.
Solution
Let’s denote the horizontal distance the box travels before hitting the ground
as xmeters. The box is initially at a height of 500 meters and falls under the
influence of gravity. The formula for the height of an object at time tunder
constant acceleration is h(t) = −1
2gt2+v0t+h0, where gis the acceleration due
to gravity (approximately 9.8 m/s2), v0is the initial vertical velocity, h0is the
initial height.
11
Since the box is initially released from rest, v0= 0. Plugging in g= 9.8 m/s2,
v0= 0, h0= 500 and h(t) = 0 into the height equation, we get:
0 = −1
2×9.8×t2+ 0 ×t+ 500
Solving for t, we find that t=q500×2
9.8.
Step 1: Calculate the time it takes for the box to hit the ground.
t=r1000
9.8≈10.2041 s
Now, to determine the horizontal distance xthe box travels before hitting the
ground, we use the equation for horizontal distance: x= horizontal velocity ×t.
The horizontal velocity of the box is equal to the horizontal speed of the
plane. Thus, x= 800 ×10.2041.
Step 2: Calculate the horizontal distance the box travels before hitting the
ground.
x= 800 ×10.2041 ≈8163.3 m
Therefore, the box travels approximately 8163.3 meters horizontally before
hitting the ground.
Question 14
Question
A rectangular garden is to be created using a wall as one side of the garden.
If the area of the garden is 300 square meters and the length of the wall is 20
meters, find the dimensions of the garden that will minimize the cost of fencing
it. Assume the cost of fencing is 10permeter.
Solution
Let xrepresent the length of the garden perpendicular to the wall and yrepre-
sent the length parallel to the wall. Therefore, the area of the garden is given
by xy and the cost of fencing the garden is 10(2x+y).
Given that the area of the garden is 300 square meters, we have:
xy = 300 ⇒y=300
x
The total cost of fencing the garden is given by C= 10(2x+y) = 20x+ 10y.
Substitute y=300
xinto the cost equation to get:
C= 20x+ 10 300
x= 20x+3000
x
12
To minimize the cost, we need to find the critical points where dC
dx = 0. So,
differentiate Cwith respect to x:
dC
dx = 20 −3000
x2
Step 1: Set dC
dx to 0 and solve for x:
20 −3000
x2= 0
3000
x2= 20
x2=3000
20 = 150
x=√150 = 10√6
Step 2: Calculate the corresponding yvalue using y=300
x:
y=300
10√6= 30√6
Therefore, the dimensions of the garden that will minimize the cost of fencing
it are 10√6 meters by 30√6 meters.
Question 15
Question
A rectangular garden has a perimeter of 80 meters. If the length of the garden
is 10 meters longer than the width, find the area of the garden.
Solution
Step 1: Let’s denote the width of the garden as xmeters. Since the length is
10 meters longer, we can express the length as x+ 10 meters.
Step 2: The perimeter of a rectangle is given by the formula P= 2L+ 2W,
where Pis the perimeter, Lis the length, and Wis the width. Substituting the
given perimeter into the formula, we have:
80 = 2(x+ 10) + 2x
Step 3: Simplifying the equation, we get:
80 = 2x+ 20 + 2x
80 = 4x+ 20
13
Step 4: Subtracting 20 from both sides gives:
60 = 4x
Step 5: Dividing both sides by 4, we find:
x= 15
Step 6: Now that we know the width is 15 meters and the length is 15+10 =
25 meters, we can calculate the area of the garden using the formula A=L×W.
A= 15 ×25
A= 375
Therefore, the area of the garden is 375 square meters.
Question 16
Question
A rectangular garden is 20 meters longer than it is wide. If the garden’s area is
288 square meters, find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Since the garden is 20 meters longer
than it is wide, the length of the garden is x+ 20 meters. We can set up the
equation for the area of the garden:
Step 1: Express the area of the garden in terms of x.
x(x+ 20) = 288
Step 2: Expand the left side of the equation.
x2+ 20x= 288
Step 3: Rearrange the equation into standard quadratic form.
x2+ 20x−288 = 0
Step 4: Solve the quadratic equation using the quadratic formula: x=
−b±√b2−4ac
2a, where a= 1, b= 20, and c=−288.
Step 5: Substitute the values of a,b, and cinto the quadratic formula and
simplify.
x=−20 ±p202−4(1)(−288)
2(1)
x=−20 ±√400 + 1152
2
14
x=−20 ±√1552
2
x=−20 ±39.37
2
Step 6: Solve for x.
x1=−20 + 39.37
2= 9.685 or x2=−20 −39.37
2=−29.685
Step 7: Since the width cannot be negative, the width of the garden is 9.685
meters. Step 8: Calculate the length by adding 20 to the width.
x+ 20 = 9.685 + 20 = 29.685
Step 9: Therefore, the dimensions of the garden are 9.685 meters by 29.685
meters.
Question 17
Question
A rectangular garden is to be created such that the length is 4 meters longer than
the width. If the area of the garden is 120 square meters, find the dimensions
of the garden.
Solution
Let’s denote the width of the garden as xmeters. Since the length is 4 meters
longer than the width, the length can be represented as x+ 4 meters. We know
that the area of a rectangle is given by the formula A= length ×width.
Step 1: Write down the equation based on the given information. The area
of the garden is given as 120 square meters. Therefore, we can write:
x(x+ 4) = 120
Step 2: Expand and simplify the equation.
x2+ 4x= 120
Step 3: Rearrange the equation into standard quadratic form.
x2+ 4x−120 = 0
Step 4: Factor the quadratic equation. To simplify the equation, we can
factor it as follows:
(x+ 12)(x−8) = 0
15
Step 5: Find the possible values of x. Setting each factor to zero gives us
two possible values for x:
x+ 12 = 0 =⇒x=−12
x−8 = 0 =⇒x= 8
Since the width cannot be negative, we take x= 8 as the width.
Step 6: Find the length of the garden. The length of the garden is x+ 4 =
8 + 4 = 12 meters.
Therefore, the dimensions of the garden are: Width = 8 meters, Length =
12 meters.
Question 18
Question
A rectangular garden is 20 meters longer than it is wide. If the garden’s area is
576 square meters, find the dimensions of the garden.
Solution
Let’s denote the width of the garden as xmeters. Since the garden is 20 meters
longer than it is wide, the length of the garden can be represented as x+ 20
meters. We are given that the area of the garden is 576 square meters.
Step 1: Write the equation for the area of the garden. The area of a
rectangle can be computed by multiplying its length by its width. Therefore,
we have:
x(x+ 20) = 576
Step 2: Solve the quadratic equation. Expanding the left side of the equa-
tion, we get:
x2+ 20x= 576
x2+ 20x−576 = 0
Step 3: Factor the quadratic equation. To solve the quadratic equation
x2+ 20x−576 = 0, we need to factor it. Since 24 ×(−24) = −576 and
24 −24 = 20, we can rewrite the equation as:
(x+ 24)(x−24) = 0
Step 4: Find the possible values of x. Setting each factor to zero gives us
the possible values of x:
x+ 24 = 0 =⇒x=−24
x−24 = 0 =⇒x= 24
16
Step 5: Check for extraneous solutions. Since the width of the garden
cannot be negative, we reject the solution x=−24. Thus, the width of the
garden is x= 24 meters.
Step 6: Find the length of the garden. The length of the garden is x+ 20 =
24 + 20 = 44 meters.
Step 7: State the dimensions of the garden. Therefore, the dimensions of
the garden are 24 meters by 44 meters.
Question 19
Question
A rectangular garden is to be constructed using a fence on three sides, with the
fourth side being a wall. The area of the garden is to be 400 square meters.
If the cost of the fencing material is
$
10 per meter and the cost of the wall is
$
100 per meter, find the dimensions of the garden that minimize the total cost
of construction.
Solution
Let the length of the garden be xmeters and the width be ymeters.
Step 1: Write the equation for the area of the garden. The area of a
rectangle is given by A=xy, and we know that A= 400 square meters.
Thus, xy = 400.
Step 2: Write the equation for the cost of construction. The cost of the
fencing will be 10($x+ 2y), since we need fencing on three sides (2 sides of
length yand 1 side of length x). The cost of the wall is 100x.
Therefore, the total cost Cis given by:
C= 10(x+ 2y) + 100x= 10x+ 20y+ 100x= 110x+ 20y
Step 3: Express yin terms of xand substitute in the total cost equation.
From Step 1, we have y=400
x. Substitute this into the total cost equation:
C= 110x+ 20 400
x
Step 4: Find the critical points by differentiating Cwith respect to xand
setting it to zero.
C′= 110 −8000
x2
Setting this derivative to zero, we get:
110 −8000
x2= 0
Solving for x, we get x= 20.
17
Step 5: Find the corresponding yvalue and check if it’s a minimum. From
Step 1, we have xy = 400, so when x= 20, y= 20. To check whether this point
is a minimum or not, we can use the second derivative test. Computing C′′, we
find:
C′′ =16000
x3
Substitute x= 20 into C′′ gives 16000
203= 4 >0, which means the point (20,20)
is a minimum.
Step 6: Final answer. So, the dimensions of the garden that minimize the
total cost of construction are 20 meters by 20 meters.
Question 20
Question
A rectangular garden has a length 5 meters longer than its width. The area of
the garden is 288 square meters. Find the dimensions of the garden.
Solution
Step 1: Let’s represent the width of the garden as xmeters. Since the length is
5 meters longer than the width, the length can be represented as x+ 5 meters.
Step 2: The area of a rectangle is given by the formula A= length ×width. In
this case, the area is 288 square meters, so we have the equation:
A= (x+ 5)x= 288
Step 3: Simplifying the equation, we get:
x2+ 5x= 288
Step 4: Rearrange the equation into standard quadratic form:
x2+ 5x−288 = 0
Step 5: We can solve this quadratic equation by factoring or using the quadratic
formula. Factoring method yields:
(x+ 24)(x−12) = 0
Step 6: Setting each factor to zero, we find the possible values for x:
x+ 24 = 0 =⇒x=−24
x−12 = 0 =⇒x= 12
Step 7: Since the width of the garden cannot be negative, we discard the negative
value. Therefore, the width of the garden is 12 meters. Step 8: The length of
the garden is x+ 5 = 12 + 5 = 17 meters. Step 9: Thus, the dimensions of the
garden are 12 meters by 17 meters.
18
Question 21
Question
A rectangular garden has a fixed perimeter of 60 feet. The length of the garden
is 6 feet longer than the width. Find the dimensions of the garden that maximize
the area.
Solution
Step 1: Let’s denote the width of the garden as wfeet. Since the length is 6
feet longer than the width, we can write the length as l=w+ 6 feet. From the
given information, we have the perimeter of the garden as 60 feet:
2w+ 2(w+ 6) = 60
2w+ 2w+ 12 = 60
4w+ 12 = 60
4w= 48
w= 12
So, the width of the garden is 12 feet and the length is 12 + 6 = 18 feet.
Step 2: The area of the rectangular garden is given by A=lw. Substituting
w= 12 and l= 18, we get:
A= 12 ×18
A= 216 ft2
Step 3: To find the dimensions that maximize the area, let’s express the area
in terms of a single variable. Since the length is l=w+ 6, we can rewrite the
area formula as:
A=w(w+ 6)
A=w2+ 6w
Step 4: To find the maximum area, we need to find the critical points of the
area function. We can do this by taking the derivative of Awith respect to w
and setting it equal to zero:
dA
dw = 2w+ 6
2w+ 6 = 0
2w=−6
w=−3
Since the width cannot be negative, there are no critical points within the
domain of possible widths.
Step 5: Since there are no critical points within the domain of possible
widths, we can conclude that the maximum area occurs at the endpoints of the
19
domain. Since the width must be positive, the only valid width is w= 12. We
already found that the corresponding length is l= 18.
Step 6: Therefore, the dimensions of the garden that maximize the area are
a width of 12 feet and a length of 18 feet, resulting in a maximum area of 216
square feet.
Question 22
Question
A farmer wants to enclose a rectangular area along a river with 100 meters
of fencing. If the side parallel to the river requires no fence, express the area
enclosed as a function of the side length perpendicular to the river.
Solution
Let the side length perpendicular to the river be xmeters. Then, the side
parallel to the river will be 100 −2xmeters.
The area of the rectangular enclosure can be expressed as the product of the
two side lengths. Therefore, the area Acan be represented as:
A=x(100 −2x)
Step 1: Expand the expression for the area.
A= 100x−2x2
Step 2: Rewrite the area function in standard quadratic form.
A=−2x2+ 100x
Step 3: This is a quadratic function. To find the maximum area, we need
to find the vertex of this parabola.
The x-coordinate of the vertex can be found using the formula x=−b
2a,
where a=−2 and b= 100.
x=−100
2(−2) = 25
Step 4: Substitute x= 25 back into the area function to find the maximum
area.
A=−2(25)2+ 100(25) = −1250 + 2500 = 1250
Step 5: Therefore, the maximum area enclosed is 1250 square meters.
20
Question 23
Question
A farmer wants to enclose a rectangular area along a river using 400 meters of
fencing. If the river forms one side of the rectangle, express the area enclosed
as a function of the width of the rectangle x.
Solution
Let the width of the rectangle be xmeters. Since the river forms one side of
the rectangle, the length of the rectangle will be 400 −2xmeters.
Step 1: The area Aof the rectangle is given by A=x(400 −2x).
Step 2: Expand the expression to simplify:
A= 400x−2x2.
Step 3: To find the maximum area that can be enclosed, we need to find
the vertex of the quadratic function A=−2x2+ 400x.
Step 4: The x-coordinate of the vertex of the parabola y=ax2+bx +cis
given by x=−b
2a. Plug in a=−2 and b= 400 to find the x-coordinate of the
vertex.
Step 5:
x=−400
2(−2) = 100
Step 6: Substitute x= 100 back into the expression for Ato find the
maximum area:
A= 400(100) −2(100)2= 40000 m2.
Step 7: Therefore, the maximum area that can be enclosed is 40000 square
meters when the width of the rectangle is 100 meters.
Question 24
Question
A rectangular garden is 20 meters longer than it is wide. If the area of the
garden is 567 square meters, find the dimensions of the garden.
Solution
Let’s denote the width of the rectangular garden as xmeters. Since the garden
is 20 meters longer than it is wide, the length can be represented as x+ 20
meters.
We are given that the area of the garden is 567 square meters. The area of
a rectangle is given by the formula A= length ×width. Substituting the given
values, we have:
x(x+ 20) = 567
21
Now, we can set up a quadratic equation by expanding the left side of the
equation:
x2+ 20x= 567
Next, we rearrange the equation in standard form:
x2+ 20x−567 = 0
To solve this quadratic equation, we can factor it or use the quadratic for-
mula. Let’s use factoring in this case:
Step 1: Factor the quadratic equation We are looking for two numbers that
multiply to -567 and add up to 20. These numbers are 27 and -21.
(x+ 27)(x−21) = 0
Step 2: Solve for x Setting each factor to zero gives us two potential solutions:
x+ 27 = 0 or x−21 = 0
x=−27 or x= 21
Since the width cannot be negative, we discard the negative solution. There-
fore, the width of the garden is 21 meters. The length can be found by adding
20:
Length = 21 + 20 = 41
Thus, the dimensions of the rectangular garden are 21 meters by 41 meters.
Question 25
Question
A rectangular garden has a fixed perimeter of 100 meters. The length of the
garden is three times its width. Find the dimensions of the garden.
Solution
Step 1: Assign variables to represent the dimensions of the garden. Let wbe
the width of the garden and lbe the length of the garden.
Step 2: Write down the given information as equations. The perimeter of a
rectangle is given by 2l+ 2w, which in this case is equal to 100 meters. Also,
we are given that the length is three times the width.
Step 3: Set up the equations based on the information given. We have:
2l+ 2w= 100
l= 3w
22
Step 4: Substitute the expression for lin terms of winto the first equation.
Substitute l= 3winto 2l+ 2w= 100:
2(3w)+2w= 100
6w+ 2w= 100
8w= 100
w= 12.5
Step 5: Find the length of the garden. Using l= 3w:
l= 3(12.5) = 37.5
Step 6: Check the answer. Let’s verify that the dimensions satisfy the
condition of the problem. The perimeter of the garden should be 100 meters:
2(37.5) + 2(12.5) = 100
75 + 25 = 100
Step 7: Write the final answer. The dimensions of the garden are width:
12.5 meters and length: 37.5 meters.
Question 26
Question
A rectangular garden has a perimeter of 50 meters. If the length of the garden
is 2 meters more than its width, find the dimensions of the garden.
Solution
Step 1: Let’s represent the width of the garden as xmeters. Since the length of
the garden is 2 meters more than the width, the length can be represented as
x+ 2 meters.
Step 2: The perimeter of a rectangle is given by 2(length + width). In this
case, the perimeter is 50 meters, so we have:
2(x+2+x) = 50
Step 3: Simplify the equation:
2(2x+ 2) = 50
4x+ 4 = 50
Step 4: Subtract 4 from both sides to isolate 4x:
4x= 46
23
Step 5: Divide by 4 to solve for x:
x=46
4= 11.5
Step 6: Calculate the length of the garden using x+ 2:
x+ 2 = 11.5 + 2 = 13.5
Step 7: Therefore, the width of the garden is 11.5 meters and the length is
13.5 meters. So, the dimensions of the garden are 11.5 meters by 13.5 meters.
Question 27
Question
A farmer wants to enclose a rectangular area alongside a river using 300 meters
of fencing. If the side along the river does not need fencing, determine the
dimensions of the rectangle that will maximize the enclosed area.
Solution
Let the dimensions of the rectangular area be xmeters and ymeters. Since the
side along the river does not need fencing, the fencing required is 2x+y= 300.
Step 1: Express one side in terms of the other side using the given perimeter.
From 2x+y= 300, we can express yin terms of x:
y= 300 −2x
Step 2: Express the area of the rectangle in terms of one variable. The area
of the rectangle is given by A=xy. Substituting y= 300 −2xinto the area
formula, we get:
A(x) = x(300 −2x) = 300x−2x2
Step 3: Find the critical points of the area function. To find the dimensions
that maximize the enclosed area, we find the critical points of the area function
A(x) by taking the derivative with respect to xand setting it equal to zero.
dA
dx = 300 −4x= 0
4x= 300
x= 75
Step 4: Determine the dimensions of the rectangle. Substitute x= 75 back
into the equation y= 300 −2x:
y= 300 −2(75)
y= 150
Therefore, the dimensions of the rectangle that will maximize the enclosed
area are 75 meters by 150 meters.
24
Question 28
Question
A rectangular garden has a fixed perimeter of 80 meters. If the area of the
garden is to be maximized, determine the dimensions of the garden.
Solution
Step 1: Let’s denote the length of the garden as xmeters and the width as y
meters.
Step 2: We are given that the perimeter of the garden is fixed at 80 meters.
Therefore, we have the equation:
2x+ 2y= 80
Step 3: Simplifying the equation, we get:
x+y= 40
Step 4: We are also given that the area of the garden is to be maximized.
The area of a rectangle is given by A=xy.
Step 5: We can express yin terms of xfrom the equation x+y= 40 as
y= 40 −x.
Step 6: Substitute y= 40 −xinto the area formula A=xy to get:
A=x(40 −x)
Step 7: Simplifying the area equation, we have:
A= 40x−x2
Step 8: To maximize the area, we need to find the critical points. Taking
the derivative of the area function:
dA
dx = 40 −2x
Step 9: Setting the derivative to zero to find the critical points:
40 −2x= 0
2x= 40
x= 20
Step 10: Substituting x= 20 back into the equation x+y= 40 to find y:
20 + y= 40
y= 20
Step 11: Therefore, the dimensions of the garden that maximize the area
are 20 meters by 20 meters.
25
Question 29
Question
A rectangular garden has a length that is 4 meters longer than its width. If the
area of the garden is 72 square meters, what are the dimensions of the garden?
Solution
Let xrepresent the width of the garden in meters. Therefore, the length of the
garden is x+ 4 meters.
Step 1: The area of a rectangle is given by the formula A= length ×width.
We are given that the area of the garden is 72 square meters. Therefore, we
have the equation:
x(x+ 4) = 72
Step 2: Simplifying the equation:
x2+ 4x= 72
x2+ 4x−72 = 0
Step 3: We can solve this quadratic equation by factoring:
x2+ 9x−8x−72 = 0
x(x+ 9) −8(x+ 9) = 0
(x−8)(x+ 9) = 0
Step 4: Setting each factor to zero gives us two possible values for the width
of the garden:
x−8 = 0 =⇒x= 8 or x+ 9 = 0 =⇒x=−9
Step 5: Since the width cannot be negative, the width of the garden is 8
meters. The length is then:
x+ 4 = 8 + 4 = 12 meters
Therefore, the dimensions of the garden are 8 meters by 12 meters.
Question 30
Question
A rectangular garden is three times as long as it is wide. If the perimeter of the
garden is 48 meters, find the dimensions of the garden.
26
Solution
Step 1: Let’s denote the width of the garden as wmeters. Since the length is
three times the width, the length of the garden is 3wmeters.
Step 2: The perimeter of a rectangle is given by the formula P= 2(l+w),
where lis the length and wis the width. Substituting the values we have,
48 = 2(3w+w)
Step 3: Simplifying the equation, we get
48 = 2(4w)
48 = 8w
Step 4: Solving for w, we find
w= 6
Step 5: Now, we can find the length of the garden by substituting w= 6
back into l= 3w:
l= 3(6) = 18
Step 6: Therefore, the dimensions of the garden are: width = 6 meters and
length = 18 meters.
Question 31
Question
A rectangular garden has a length 5 meters longer than its width. If the area
of the garden is 144 square meters, find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Then, the length of the garden is x+5
meters.
Step 1: Write an equation for the area of the garden in terms of x:
x(x+ 5) = 144
Step 2: Expand the left side of the equation:
x2+ 5x= 144
Step 3: Rearrange the equation to form a quadratic equation in standard
form:
x2+ 5x−144 = 0
27
Step 4: Solve the quadratic equation using the quadratic formula: The
quadratic formula is given by x=−b±√b2−4ac
2a, where a= 1, b= 5, and
c=−144.
x=−5±p52−4(1)(−144)
2(1)
x=−5±√25 + 576
2
x=−5±√601
2
x=−5±√601
2
Step 5: Based on the problem, we should choose the positive value for x
since the dimensions must be positive. Therefore,
x=−5 + √601
2≈8.89
Step 6: Calculate the length by adding 5 to x:
x+ 5 ≈8.89 + 5 = 13.89
So, the dimensions of the garden are approximately 8.89 meters by 13.89
meters.
Question 32
Question
A rectangular garden is 20 meters longer than it is wide. If the area of the
garden is 525 square meters, find the dimensions of the garden.
Solution
Step 1: Let’s denote the width of the garden as xmeters. Since the garden is 20
meters longer than it is wide, the length of the garden would be x+ 20 meters.
Step 2: The area of a rectangle is given by the formula A= length ×width.
Substituting the given values into this formula, we have:
x(x+ 20) = 525
Step 3: Expanding the left side of the equation, we get:
x2+ 20x= 525
Step 4: Rearranging this equation into standard quadratic form, we have:
28
x2+ 20x−525 = 0
Step 5: To solve this quadratic equation, we can factorize it. The factors of
−525 that add up to 20 are 35 and −15. Therefore, we can rewrite the equation
as:
(x+ 35)(x−15) = 0
Step 6: Setting each factor to zero gives us two possible values for the width:
x+ 35 = 0 =⇒x=−35 (extraneous solution)
x−15 = 0 =⇒x= 15
Step 7: Since the width cannot be negative, the width of the garden is x= 15
meters.
Step 8: Substituting x= 15 back into the expression for the length (x+ 20)
gives us the length of the garden:
15 + 20 = 35
Step 9: Therefore, the dimensions of the garden are 15 meters by 35 meters.
Question 33
Question
A rectangular garden is twice as long as it is wide. If the area of the garden is
192 square meters, find the dimensions of the garden.
Solution
Step 1: Let’s denote the width of the garden as wmeters. Since the length is
twice the width, we can represent the length as 2wmeters.
Step 2: The area of a rectangle is given by the formula A= length ×width.
Substituting the given values into the formula, we have:
192 = 2w×w
Step 3: Simplifying the equation, we get:
192 = 2w2
Step 4: Divide both sides of the equation by 2 to solve for w2:
w2= 96
Step 5: Taking the square root of both sides, we find:
w=√96 = 4√6
29
Step 6: Therefore, the width of the garden is 4√6 meters. Since the length
is twice the width, the length is:
2w= 2 ×4√6=8√6
Step 7: Thus, the dimensions of the garden are width = 4√6 meters and
length = 8√6 meters.
Question 34
Question
A rectangular garden with an area of 2400 square meters is to be surrounded
by a walkway of uniform width. If the walking area is to be 10 meters wide,
find the dimensions of the entire garden.
Solution
Let the width of the garden be xmeters. Since the area of the garden is given
by the product of its length and width, we have:
Step 1: Setting up the equation to represent the area of the garden:
x(x+ 20) = 2400
Step 2: Expand the left side of the equation:
x2+ 20x= 2400
Step 3: Rearrange the equation into a standard quadratic form:
x2+ 20x−2400 = 0
Step 4: Factor the quadratic equation:
(x+ 60)(x−40) = 0
Step 5: Solve for the possible values of xby setting each factor to zero:
x+ 60 = 0 =⇒x=−60 or x−40 = 0 =⇒x= 40
Step 6: Since the width cannot be negative, the width of the garden is 40
meters. Now, we can find the length of the garden:
Length = x+ 20 = 40 + 20 = 60
Therefore, the dimensions of the entire garden are 40 meters by 60 meters.
30
Question 35
Question
A rectangular garden is 20 meters long and 15 meters wide. A path of uniform
width is to be built around the garden. If the total area of the garden and the
path is 600 square meters, find the width of the path.
Solution
Let the width of the path be xmeters.
Step 1: The total area of the garden and the path is the sum of the area of
the garden and the area of the path. Set up an equation based on this fact.
Total area = Area of garden + Area of path
Step 2: Calculate the area of the garden.
Area of garden = length ×width = 20 ×15 square meters
Area of garden = 300 square meters
Step 3: Calculate the dimensions of the entire region including the path.
The length of the entire region is 20 + 2xmeters (added width on both sides)
and the width is 15 + 2xmeters. Thus, the area of the entire region including
the path is
Area of region = (20 + 2x)(15 + 2x) square meters
Step 4: Form an equation using the total area given.
600 = 300 + (20 + 2x)(15 + 2x)
Step 5: Solve the equation by expanding and simplifying.
600 = 300 + 300 + 50x+ 40x+ 4x2
4x2+ 90x−300 = 0
Step 6: Factor the quadratic equation.
4x2+ 90x−300 = 0
x2+ 22.5x−75 = 0
(x+ 25)(x−3) = 0
Step 7: Find the possible solutions for x.
x=−25 (extraneous solution) or x= 3
Step 8: Check the solution. As the width cannot be negative, the width of
the path is 3 meters.
31
Question 2
Question
A rectangular garden has a fixed perimeter of 60 meters. The length of the
garden is 4 meters longer than its width. Find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Then, the length of the garden is x+4
meters.
Step 1: Write an equation based on the information given. The perimeter
of a rectangle is given by 2 ×(length + width). Since the perimeter is 60 meters,
we have:
2×(x+ (x+ 4)) = 60
Step 2: Solve the equation for x.
2×(2x+ 4) = 60
4x+ 8 = 60
4x= 52
x= 13
Step 3: Find the dimensions of the garden. The width, x, is 13 meters.
The length, x+ 4, is 13 + 4 = 17 meters.
Step 4: Check the answer. The perimeter of the garden is 2 ×(13 + 17) =
2×30 = 60 meters, which matches the given fixed perimeter. Therefore, the
dimensions of the garden are width: 13 meters and length: 17 meters.
Question 3
Question
A rectangular garden has a fixed perimeter of 40 meters. If the area of the
garden is to be maximized, what should the dimensions of the garden be?
Solution
Step 1: Let xand yrepresent the dimensions of the garden. Step 2: Write the
equation for the perimeter of the garden in terms of xand y. Step 3: Write
the equation for the area of the garden in terms of xand y. Step 4: Express
yin terms of xin the equation for the area. Step 5: Find the critical points
by taking the derivative of the area equation with respect to xand setting it
equal to 0. Step 6: Determine if it is a maximum or minimum point by using
the second derivative test. Step 7: Calculate the dimensions of the garden at
the maximum area.
2
Let’s find the dimensions of the garden that maximize the area.
Step 1: Let xand yrepresent the dimensions of the garden.
Step 2: The perimeter of a rectangle is given by 2x+2y= 40, or simplifying,
x+y= 20.
Step 3: The area of a rectangle is given by A=xy.
Step 4: Since y= 20 −x, substitute this into the area equation to get
A(x) = x(20 −x) = 20x−x2.
Step 5: Find the critical points by taking the derivative of A(x) and setting
it equal to 0:
dA
dx = 20 −2x= 0
20 = 2x
x= 10
Step 6: Use the second derivative test to check if x= 10 is a maximum:
d2A
dx2=−2<0
Since the second derivative is negative, the point is a maximum.
Step 7: Now that we have x= 10, substitute it back into y= 20 −x:
y= 20 −10
y= 10
Therefore, the dimensions of the garden with the maximum area are 10
meters by 10 meters.
Question 4
Question
A rectangular garden is to be created using a wall as one of its sides. Three
sides of the garden are to be enclosed with a fence, while the fourth side is the
existing wall. If the area of the garden is 200 square meters and the length of
the garden is 10 meters longer than its width, find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Then the length of the garden is x+10
meters.
The area of a rectangle is given by the formula A= width ×length. We are
given that the area of the garden is 200 square meters, so we have the equation:
x(x+ 10) = 200
3
Expanding and simplifying this equation, we get:
x2+ 10x−200 = 0
To solve this quadratic equation, we can either factorize it or use the quadratic
formula. Let’s use the quadratic formula: The quadratic formula states that for
an equation of the form ax2+bx +c= 0, the solutions are given by:
x=−b±√b2−4ac
2a
In this case, a= 1, b= 10, and c=−200. Plugging these values into the
formula, we get:
x=−10 ±p102−4(1)(−200)
2(1)
x=−10 ±√100 + 800
2
x=−10 ±√900
2
Therefore, we have two possible solutions for the width of the garden:
x=−10 + 30
2=20
2= 10
x=−10 −30
2=−40
2=−20
Since the width cannot be negative, we take x= 10.
Thus, the width of the garden is 10 meters and the length is 10 + 10 = 20
meters. So, the dimensions of the garden are 10 meters by 20 meters.
Question 5
Question
A rectangular garden has a length that is 6 meters longer than its width. If the
area of the garden is 300 square meters, find the dimensions of the garden.
Solution
1. Let’s denote the width of the garden as wmeters. We know that the length of
the garden is 6 meters longer, so the length can be represented as w+ 6 meters.
2. The area of a rectangle is given by the formula A= length ×width.
Substituting our expressions for length and width into this formula, we have:
w(w+ 6) = 300
4
3. Now, we can simplify the equation by expanding the product on the left
side:
w2+ 6w= 300
4. Rearranging the equation to set it equal to zero, we get a quadratic
equation:
w2+ 6w−300 = 0
5. To solve this quadratic equation, we can factor it. We look for two
numbers that multiply to -300 and add to 6. These numbers are 20 and -15.
So, we can rewrite the equation as:
w2+ 20w−15w−300 = 0
6. Factor by grouping:
w(w+ 20) −15(w+ 20) = 0
(w−15)(w+ 20) = 0
7. By setting each factor to zero, we find the possible values for the width:
w−15 = 0 or w+ 20 = 0
w= 15 or w=−20
8. Since the width cannot be negative, we discard w=−20 and conclude
that the width is w= 15 meters.
9. Using this width, we can find the length:
length = w+ 6 = 15 + 6 = 21
10. Therefore, the dimensions of the garden are a width of 15 meters and a
length of 21 meters.
Question 6
Question
A rectangular garden has a perimeter of 100 meters. The length of the garden
is 10 meters longer than its width. Find the dimensions of the garden.
Solution
Let’s denote the width of the garden as xmeters. Since the length is 10 meters
longer than the width, the length can be represented as x+ 10 meters.
Step 1: Write down the equation based on the given information. The
perimeter of a rectangle is given by 2×(length + width). From the problem, we
5
know that the perimeter is 100 meters. Therefore, the equation can be written
as:
2(x+x+ 10) = 100
Step 2: Simplify the equation.
4x+ 20 = 100
4x= 80
x= 20
Step 3: Find the length. The length of the garden is x+ 10 meters.
x+ 10 = 20 + 10 = 30
Step 4: State the dimensions of the garden. Therefore, the width of the
garden is 20 meters and the length is 30 meters. The dimensions of the garden
are 20 meters by 30 meters.
Question 7
Question
A rectangular flower garden is 12 meters long and 8 meters wide. The garden is
to be enlarged by increasing both the length and the width by the same amount.
If the area of the new garden is 208 square meters more than the area of the
original garden, find the dimensions of the new garden.
Solution
Let xbe the amount by which both the length and the width are increased.
Step 1: Determine the area of the original garden. The area of the original
garden is given by:
Areaoriginal = length ×width = 12 ×8 = 96 sq meters
Step 2: Determine the area of the new garden. After increasing both the
length and the width by x, the new dimensions are 12 + xmeters and 8 + x
meters. The area of the new garden is then:
Areanew = (12 + x)(8 + x) = 96 + 20x+x2
Step 3: Set up an equation based on the given information. Since the new
garden is 208 square meters larger than the original garden, we have:
Areanew = Areaoriginal + 208
96 + 20x+x2= 96 + 208
6
Step 4: Solve the equation.
x2+ 20x−208 = 0
Factoring the quadratic equation, we get:
(x+ 26)(x−8) = 0
So, x=−26 or x= 8. Since the dimensions cannot be negative, the amount by
which the garden is enlarged is 8 meters.
Step 5: Determine the dimensions of the new garden. The dimensions of
the new garden are: Length: 12 + 8 = 20 meters Width: 8 + 8 = 16 meters
Therefore, the new garden is 20 meters long and 16 meters wide.
Question 8
Question
A company manufactures and sells a certain product. The revenue from selling
xunits of the product is given by the function R(x) = −2x2+100x. The cost of
producing xunits of the product is given by the function C(x)=0.5x2+40x+10.
Find:
1. The profit function P(x).
2. The number of units that must be sold to maximize profit.
3. The maximum profit possible.
Solution
1. To find the profit function P(x), we subtract the cost function from the
revenue function:
P(x) = R(x)−C(x)=(−2x2+ 100x)−(0.5x2+ 40x+ 10)
Simplify to get:
P(x) = −2x2+ 100x−0.5x2−40x−10 = −2.5x2+ 60x−10
Therefore, the profit function is P(x) = −2.5x2+ 60x−10.
2. To find the number of units that must be sold to maximize profit, we first
need to find the critical points of the profit function P(x). The profit function is
a quadratic function, and its maximum will occur at the vertex of the parabola.
The x-coordinate of the vertex of a parabola in the form ax2+bx +cis given
by −b
2a. In this case, a=−2.5 and b= 60. Therefore, the number of units to
be sold to maximize profit is:
x=−60
2(−2.5) =−60
−5= 12
7
.
3. To find the maximum profit possible, substitute x= 12 into the profit
function:
P(12) = −2.5(12)2+ 60(12) −10 = −360 + 720 −10 = 350
Therefore, the maximum profit possible is 350.
Question 9
Question
A photographer is hired to take wedding photos at a rate of xdollars per
photo. The photographer takes a total of 50 photos during the wedding. If
the photographer is paid a total of $500 for the photos, find the rate per photo.
Solution
Let xbe the rate per photo in dollars.
Step 1: Write an equation using the given information. The total pay for 50
photos is given by the product of the rate per photo and the number of photos
taken. Since the total pay is $500, we have:
50x= 500
Step 2: Solve the equation to find the rate per photo. Divide both sides of
the equation by 50:
x=500
50
x= 10
Step 3: Answer The rate per photo is $10.
Question 10
Question
A rectangular garden is to be created using 60 meters of fencing. If one side of
the garden is against a wall, find the dimensions of the garden that maximize
the area.
Solution
Let xbe the length of the side of the garden against the wall and ybe the width
of the garden.
Step 1: Express the total perimeter in terms of xand y. The perimeter of
the garden is given as 60 meters, so we have the equation: 2x+y= 60.
8
Step 2: Express the area of the garden in terms of xand y. The area of
the garden is given by A=xy.
Step 3: Express yin terms of x. From the perimeter equation, we can
express yas y= 60 −2x.
Step 4: Substitute y= 60−2xinto the area formula. Substitute y= 60−2x
into the area formula A=xy to get A=x(60 −2x), which simplifies to A=
60x−2x2.
Step 5: Find the critical points by taking the derivative of the area function.
To find the dimensions that maximize the area, we need to find the critical points
by taking the derivative of the area function: dA
dx = 60 −4x.
Step 6: Set the derivative equal to zero and solve for x. Set dA
dx = 60−4x= 0
and solve for x: 60 = 4x=⇒x= 15.
Step 7: Find the corresponding value of y. To find the corresponding value
of y, substitute x= 15 into y= 60 −2x:y= 60 −2(15) = 60 −30 = 30.
Step 8: Check the endpoints. Check the endpoints of the possible interval
to ensure the maximum is at x= 15.
Step 9: State the dimensions that maximize the area. Therefore, the di-
mensions that maximize the area are x= 15 meters (length) and y= 30 meters
(width).
Question 11
Question
A rectangular playground has a length that is 10 meters longer than its width.
If the area of the playground is 192 square meters, find the dimensions of the
playground.
Solution
Let’s denote the width of the playground as xmeters. Since the length is 10
meters longer than the width, we can express the length as x+ 10 meters.
We know that the area of a rectangle is given by the formula A= length ×
width. For this playground, the area is given as 192 square meters:
x(x+ 10) = 192
Expanding the left side of the equation:
x2+ 10x= 192
Rearranging the equation into standard form:
x2+ 10x−192 = 0
Next, we can solve this quadratic equation using the quadratic formula:
9
x=−b±√b2−4ac
2a
where a= 1, b= 10, and c=−192.
Plugging these values into the quadratic formula:
x=−10 ±√102−4×1× −192
2×1
x=−10 ±√100 + 768
2
x=−10 ±√868
2
x=−10 ±2√217
2
x=−5±√217
Therefore, the possible values for the width are −5 + √217 and −5−√217.
Since the width cannot be negative, the correct value for the width is −5+√217.
Now, we can find the length by substituting the width back into the expres-
sion x+ 10:
Length = (−5 + √217) + 10
Length = 5 + √217
So, the width of the playground is approximately 1.78 meters and the length
is approximately 15.78 meters.
Question 12
Question
A rectangular garden is four times as long as it is wide. If the area of the garden
is 100 square meters, find the dimensions of the garden.
Solution
Let’s denote the width of the garden as wmeters. Since the length of the garden
is four times the width, the length can be represented as 4wmeters. We are
given that the area of the garden is 100 square meters.
Step 1: Write the equation for the area of the garden in terms of w. The
area of a rectangle is given by the formula A= length ×width. Substituting
the given values, we have:
100 = 4w×w
10
Step 2: Simplify the equation to form a quadratic equation.
100 = 4w2
Step 3: Rewrite the equation in standard quadratic form.
4w2−100 = 0
Step 4: Solve the quadratic equation by factoring or using the quadratic
formula. We can factor out a 4 from the equation:
4(w2−25) = 0
Step 5: Solve for wby setting each factor to zero. We have two possible
solutions:
w2−25 = 0
Step 6: Write down the two possible solutions for w.
w2−25 = 0
w2= 25
w=±5
Since the width cannot be negative, the width of the garden is 5 meters.
Step 7: Find the length of the garden. The length of the garden is 4 times
the width, which is:
4×5 = 20
Step 8: Write the final answer. The dimensions of the garden are width =
5 meters and length = 20 meters.
Question 13
Question
A cargo plane flying at a constant speed releases a box of supplies at a height of
500 meters above the ground. The horizontal distance between the plane and
the point directly below the box when the box is released is 800 meters. Find
the horizontal distance the box travels before hitting the ground.
Solution
Let’s denote the horizontal distance the box travels before hitting the ground
as xmeters. The box is initially at a height of 500 meters and falls under the
influence of gravity. The formula for the height of an object at time tunder
constant acceleration is h(t) = −1
2gt2+v0t+h0, where gis the acceleration due
to gravity (approximately 9.8 m/s2), v0is the initial vertical velocity, h0is the
initial height.
11
Since the box is initially released from rest, v0= 0. Plugging in g= 9.8 m/s2,
v0= 0, h0= 500 and h(t) = 0 into the height equation, we get:
0 = −1
2×9.8×t2+ 0 ×t+ 500
Solving for t, we find that t=q500×2
9.8.
Step 1: Calculate the time it takes for the box to hit the ground.
t=r1000
9.8≈10.2041 s
Now, to determine the horizontal distance xthe box travels before hitting the
ground, we use the equation for horizontal distance: x= horizontal velocity ×t.
The horizontal velocity of the box is equal to the horizontal speed of the
plane. Thus, x= 800 ×10.2041.
Step 2: Calculate the horizontal distance the box travels before hitting the
ground.
x= 800 ×10.2041 ≈8163.3 m
Therefore, the box travels approximately 8163.3 meters horizontally before
hitting the ground.
Question 14
Question
A rectangular garden is to be created using a wall as one side of the garden.
If the area of the garden is 300 square meters and the length of the wall is 20
meters, find the dimensions of the garden that will minimize the cost of fencing
it. Assume the cost of fencing is 10permeter.
Solution
Let xrepresent the length of the garden perpendicular to the wall and yrepre-
sent the length parallel to the wall. Therefore, the area of the garden is given
by xy and the cost of fencing the garden is 10(2x+y).
Given that the area of the garden is 300 square meters, we have:
xy = 300 ⇒y=300
x
The total cost of fencing the garden is given by C= 10(2x+y) = 20x+ 10y.
Substitute y=300
xinto the cost equation to get:
C= 20x+ 10 300
x= 20x+3000
x
12
To minimize the cost, we need to find the critical points where dC
dx = 0. So,
differentiate Cwith respect to x:
dC
dx = 20 −3000
x2
Step 1: Set dC
dx to 0 and solve for x:
20 −3000
x2= 0
3000
x2= 20
x2=3000
20 = 150
x=√150 = 10√6
Step 2: Calculate the corresponding yvalue using y=300
x:
y=300
10√6= 30√6
Therefore, the dimensions of the garden that will minimize the cost of fencing
it are 10√6 meters by 30√6 meters.
Question 15
Question
A rectangular garden has a perimeter of 80 meters. If the length of the garden
is 10 meters longer than the width, find the area of the garden.
Solution
Step 1: Let’s denote the width of the garden as xmeters. Since the length is
10 meters longer, we can express the length as x+ 10 meters.
Step 2: The perimeter of a rectangle is given by the formula P= 2L+ 2W,
where Pis the perimeter, Lis the length, and Wis the width. Substituting the
given perimeter into the formula, we have:
80 = 2(x+ 10) + 2x
Step 3: Simplifying the equation, we get:
80 = 2x+ 20 + 2x
80 = 4x+ 20
13
Step 4: Subtracting 20 from both sides gives:
60 = 4x
Step 5: Dividing both sides by 4, we find:
x= 15
Step 6: Now that we know the width is 15 meters and the length is 15+10 =
25 meters, we can calculate the area of the garden using the formula A=L×W.
A= 15 ×25
A= 375
Therefore, the area of the garden is 375 square meters.
Question 16
Question
A rectangular garden is 20 meters longer than it is wide. If the garden’s area is
288 square meters, find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Since the garden is 20 meters longer
than it is wide, the length of the garden is x+ 20 meters. We can set up the
equation for the area of the garden:
Step 1: Express the area of the garden in terms of x.
x(x+ 20) = 288
Step 2: Expand the left side of the equation.
x2+ 20x= 288
Step 3: Rearrange the equation into standard quadratic form.
x2+ 20x−288 = 0
Step 4: Solve the quadratic equation using the quadratic formula: x=
−b±√b2−4ac
2a, where a= 1, b= 20, and c=−288.
Step 5: Substitute the values of a,b, and cinto the quadratic formula and
simplify.
x=−20 ±p202−4(1)(−288)
2(1)
x=−20 ±√400 + 1152
2
14
x=−20 ±√1552
2
x=−20 ±39.37
2
Step 6: Solve for x.
x1=−20 + 39.37
2= 9.685 or x2=−20 −39.37
2=−29.685
Step 7: Since the width cannot be negative, the width of the garden is 9.685
meters. Step 8: Calculate the length by adding 20 to the width.
x+ 20 = 9.685 + 20 = 29.685
Step 9: Therefore, the dimensions of the garden are 9.685 meters by 29.685
meters.
Question 17
Question
A rectangular garden is to be created such that the length is 4 meters longer than
the width. If the area of the garden is 120 square meters, find the dimensions
of the garden.
Solution
Let’s denote the width of the garden as xmeters. Since the length is 4 meters
longer than the width, the length can be represented as x+ 4 meters. We know
that the area of a rectangle is given by the formula A= length ×width.
Step 1: Write down the equation based on the given information. The area
of the garden is given as 120 square meters. Therefore, we can write:
x(x+ 4) = 120
Step 2: Expand and simplify the equation.
x2+ 4x= 120
Step 3: Rearrange the equation into standard quadratic form.
x2+ 4x−120 = 0
Step 4: Factor the quadratic equation. To simplify the equation, we can
factor it as follows:
(x+ 12)(x−8) = 0
15
Step 5: Find the possible values of x. Setting each factor to zero gives us
two possible values for x:
x+ 12 = 0 =⇒x=−12
x−8 = 0 =⇒x= 8
Since the width cannot be negative, we take x= 8 as the width.
Step 6: Find the length of the garden. The length of the garden is x+ 4 =
8 + 4 = 12 meters.
Therefore, the dimensions of the garden are: Width = 8 meters, Length =
12 meters.
Question 18
Question
A rectangular garden is 20 meters longer than it is wide. If the garden’s area is
576 square meters, find the dimensions of the garden.
Solution
Let’s denote the width of the garden as xmeters. Since the garden is 20 meters
longer than it is wide, the length of the garden can be represented as x+ 20
meters. We are given that the area of the garden is 576 square meters.
Step 1: Write the equation for the area of the garden. The area of a
rectangle can be computed by multiplying its length by its width. Therefore,
we have:
x(x+ 20) = 576
Step 2: Solve the quadratic equation. Expanding the left side of the equa-
tion, we get:
x2+ 20x= 576
x2+ 20x−576 = 0
Step 3: Factor the quadratic equation. To solve the quadratic equation
x2+ 20x−576 = 0, we need to factor it. Since 24 ×(−24) = −576 and
24 −24 = 20, we can rewrite the equation as:
(x+ 24)(x−24) = 0
Step 4: Find the possible values of x. Setting each factor to zero gives us
the possible values of x:
x+ 24 = 0 =⇒x=−24
x−24 = 0 =⇒x= 24
16
Step 5: Check for extraneous solutions. Since the width of the garden
cannot be negative, we reject the solution x=−24. Thus, the width of the
garden is x= 24 meters.
Step 6: Find the length of the garden. The length of the garden is x+ 20 =
24 + 20 = 44 meters.
Step 7: State the dimensions of the garden. Therefore, the dimensions of
the garden are 24 meters by 44 meters.
Question 19
Question
A rectangular garden is to be constructed using a fence on three sides, with the
fourth side being a wall. The area of the garden is to be 400 square meters.
If the cost of the fencing material is
$
10 per meter and the cost of the wall is
$
100 per meter, find the dimensions of the garden that minimize the total cost
of construction.
Solution
Let the length of the garden be xmeters and the width be ymeters.
Step 1: Write the equation for the area of the garden. The area of a
rectangle is given by A=xy, and we know that A= 400 square meters.
Thus, xy = 400.
Step 2: Write the equation for the cost of construction. The cost of the
fencing will be 10($x+ 2y), since we need fencing on three sides (2 sides of
length yand 1 side of length x). The cost of the wall is 100x.
Therefore, the total cost Cis given by:
C= 10(x+ 2y) + 100x= 10x+ 20y+ 100x= 110x+ 20y
Step 3: Express yin terms of xand substitute in the total cost equation.
From Step 1, we have y=400
x. Substitute this into the total cost equation:
C= 110x+ 20 400
x
Step 4: Find the critical points by differentiating Cwith respect to xand
setting it to zero.
C′= 110 −8000
x2
Setting this derivative to zero, we get:
110 −8000
x2= 0
Solving for x, we get x= 20.
17
Step 5: Find the corresponding yvalue and check if it’s a minimum. From
Step 1, we have xy = 400, so when x= 20, y= 20. To check whether this point
is a minimum or not, we can use the second derivative test. Computing C′′, we
find:
C′′ =16000
x3
Substitute x= 20 into C′′ gives 16000
203= 4 >0, which means the point (20,20)
is a minimum.
Step 6: Final answer. So, the dimensions of the garden that minimize the
total cost of construction are 20 meters by 20 meters.
Question 20
Question
A rectangular garden has a length 5 meters longer than its width. The area of
the garden is 288 square meters. Find the dimensions of the garden.
Solution
Step 1: Let’s represent the width of the garden as xmeters. Since the length is
5 meters longer than the width, the length can be represented as x+ 5 meters.
Step 2: The area of a rectangle is given by the formula A= length ×width. In
this case, the area is 288 square meters, so we have the equation:
A= (x+ 5)x= 288
Step 3: Simplifying the equation, we get:
x2+ 5x= 288
Step 4: Rearrange the equation into standard quadratic form:
x2+ 5x−288 = 0
Step 5: We can solve this quadratic equation by factoring or using the quadratic
formula. Factoring method yields:
(x+ 24)(x−12) = 0
Step 6: Setting each factor to zero, we find the possible values for x:
x+ 24 = 0 =⇒x=−24
x−12 = 0 =⇒x= 12
Step 7: Since the width of the garden cannot be negative, we discard the negative
value. Therefore, the width of the garden is 12 meters. Step 8: The length of
the garden is x+ 5 = 12 + 5 = 17 meters. Step 9: Thus, the dimensions of the
garden are 12 meters by 17 meters.
18
Question 21
Question
A rectangular garden has a fixed perimeter of 60 feet. The length of the garden
is 6 feet longer than the width. Find the dimensions of the garden that maximize
the area.
Solution
Step 1: Let’s denote the width of the garden as wfeet. Since the length is 6
feet longer than the width, we can write the length as l=w+ 6 feet. From the
given information, we have the perimeter of the garden as 60 feet:
2w+ 2(w+ 6) = 60
2w+ 2w+ 12 = 60
4w+ 12 = 60
4w= 48
w= 12
So, the width of the garden is 12 feet and the length is 12 + 6 = 18 feet.
Step 2: The area of the rectangular garden is given by A=lw. Substituting
w= 12 and l= 18, we get:
A= 12 ×18
A= 216 ft2
Step 3: To find the dimensions that maximize the area, let’s express the area
in terms of a single variable. Since the length is l=w+ 6, we can rewrite the
area formula as:
A=w(w+ 6)
A=w2+ 6w
Step 4: To find the maximum area, we need to find the critical points of the
area function. We can do this by taking the derivative of Awith respect to w
and setting it equal to zero:
dA
dw = 2w+ 6
2w+ 6 = 0
2w=−6
w=−3
Since the width cannot be negative, there are no critical points within the
domain of possible widths.
Step 5: Since there are no critical points within the domain of possible
widths, we can conclude that the maximum area occurs at the endpoints of the
19
domain. Since the width must be positive, the only valid width is w= 12. We
already found that the corresponding length is l= 18.
Step 6: Therefore, the dimensions of the garden that maximize the area are
a width of 12 feet and a length of 18 feet, resulting in a maximum area of 216
square feet.
Question 22
Question
A farmer wants to enclose a rectangular area along a river with 100 meters
of fencing. If the side parallel to the river requires no fence, express the area
enclosed as a function of the side length perpendicular to the river.
Solution
Let the side length perpendicular to the river be xmeters. Then, the side
parallel to the river will be 100 −2xmeters.
The area of the rectangular enclosure can be expressed as the product of the
two side lengths. Therefore, the area Acan be represented as:
A=x(100 −2x)
Step 1: Expand the expression for the area.
A= 100x−2x2
Step 2: Rewrite the area function in standard quadratic form.
A=−2x2+ 100x
Step 3: This is a quadratic function. To find the maximum area, we need
to find the vertex of this parabola.
The x-coordinate of the vertex can be found using the formula x=−b
2a,
where a=−2 and b= 100.
x=−100
2(−2) = 25
Step 4: Substitute x= 25 back into the area function to find the maximum
area.
A=−2(25)2+ 100(25) = −1250 + 2500 = 1250
Step 5: Therefore, the maximum area enclosed is 1250 square meters.
20
Question 23
Question
A farmer wants to enclose a rectangular area along a river using 400 meters of
fencing. If the river forms one side of the rectangle, express the area enclosed
as a function of the width of the rectangle x.
Solution
Let the width of the rectangle be xmeters. Since the river forms one side of
the rectangle, the length of the rectangle will be 400 −2xmeters.
Step 1: The area Aof the rectangle is given by A=x(400 −2x).
Step 2: Expand the expression to simplify:
A= 400x−2x2.
Step 3: To find the maximum area that can be enclosed, we need to find
the vertex of the quadratic function A=−2x2+ 400x.
Step 4: The x-coordinate of the vertex of the parabola y=ax2+bx +cis
given by x=−b
2a. Plug in a=−2 and b= 400 to find the x-coordinate of the
vertex.
Step 5:
x=−400
2(−2) = 100
Step 6: Substitute x= 100 back into the expression for Ato find the
maximum area:
A= 400(100) −2(100)2= 40000 m2.
Step 7: Therefore, the maximum area that can be enclosed is 40000 square
meters when the width of the rectangle is 100 meters.
Question 24
Question
A rectangular garden is 20 meters longer than it is wide. If the area of the
garden is 567 square meters, find the dimensions of the garden.
Solution
Let’s denote the width of the rectangular garden as xmeters. Since the garden
is 20 meters longer than it is wide, the length can be represented as x+ 20
meters.
We are given that the area of the garden is 567 square meters. The area of
a rectangle is given by the formula A= length ×width. Substituting the given
values, we have:
x(x+ 20) = 567
21
Now, we can set up a quadratic equation by expanding the left side of the
equation:
x2+ 20x= 567
Next, we rearrange the equation in standard form:
x2+ 20x−567 = 0
To solve this quadratic equation, we can factor it or use the quadratic for-
mula. Let’s use factoring in this case:
Step 1: Factor the quadratic equation We are looking for two numbers that
multiply to -567 and add up to 20. These numbers are 27 and -21.
(x+ 27)(x−21) = 0
Step 2: Solve for x Setting each factor to zero gives us two potential solutions:
x+ 27 = 0 or x−21 = 0
x=−27 or x= 21
Since the width cannot be negative, we discard the negative solution. There-
fore, the width of the garden is 21 meters. The length can be found by adding
20:
Length = 21 + 20 = 41
Thus, the dimensions of the rectangular garden are 21 meters by 41 meters.
Question 25
Question
A rectangular garden has a fixed perimeter of 100 meters. The length of the
garden is three times its width. Find the dimensions of the garden.
Solution
Step 1: Assign variables to represent the dimensions of the garden. Let wbe
the width of the garden and lbe the length of the garden.
Step 2: Write down the given information as equations. The perimeter of a
rectangle is given by 2l+ 2w, which in this case is equal to 100 meters. Also,
we are given that the length is three times the width.
Step 3: Set up the equations based on the information given. We have:
2l+ 2w= 100
l= 3w
22
Step 4: Substitute the expression for lin terms of winto the first equation.
Substitute l= 3winto 2l+ 2w= 100:
2(3w)+2w= 100
6w+ 2w= 100
8w= 100
w= 12.5
Step 5: Find the length of the garden. Using l= 3w:
l= 3(12.5) = 37.5
Step 6: Check the answer. Let’s verify that the dimensions satisfy the
condition of the problem. The perimeter of the garden should be 100 meters:
2(37.5) + 2(12.5) = 100
75 + 25 = 100
Step 7: Write the final answer. The dimensions of the garden are width:
12.5 meters and length: 37.5 meters.
Question 26
Question
A rectangular garden has a perimeter of 50 meters. If the length of the garden
is 2 meters more than its width, find the dimensions of the garden.
Solution
Step 1: Let’s represent the width of the garden as xmeters. Since the length of
the garden is 2 meters more than the width, the length can be represented as
x+ 2 meters.
Step 2: The perimeter of a rectangle is given by 2(length + width). In this
case, the perimeter is 50 meters, so we have:
2(x+2+x) = 50
Step 3: Simplify the equation:
2(2x+ 2) = 50
4x+ 4 = 50
Step 4: Subtract 4 from both sides to isolate 4x:
4x= 46
23
Step 5: Divide by 4 to solve for x:
x=46
4= 11.5
Step 6: Calculate the length of the garden using x+ 2:
x+ 2 = 11.5 + 2 = 13.5
Step 7: Therefore, the width of the garden is 11.5 meters and the length is
13.5 meters. So, the dimensions of the garden are 11.5 meters by 13.5 meters.
Question 27
Question
A farmer wants to enclose a rectangular area alongside a river using 300 meters
of fencing. If the side along the river does not need fencing, determine the
dimensions of the rectangle that will maximize the enclosed area.
Solution
Let the dimensions of the rectangular area be xmeters and ymeters. Since the
side along the river does not need fencing, the fencing required is 2x+y= 300.
Step 1: Express one side in terms of the other side using the given perimeter.
From 2x+y= 300, we can express yin terms of x:
y= 300 −2x
Step 2: Express the area of the rectangle in terms of one variable. The area
of the rectangle is given by A=xy. Substituting y= 300 −2xinto the area
formula, we get:
A(x) = x(300 −2x) = 300x−2x2
Step 3: Find the critical points of the area function. To find the dimensions
that maximize the enclosed area, we find the critical points of the area function
A(x) by taking the derivative with respect to xand setting it equal to zero.
dA
dx = 300 −4x= 0
4x= 300
x= 75
Step 4: Determine the dimensions of the rectangle. Substitute x= 75 back
into the equation y= 300 −2x:
y= 300 −2(75)
y= 150
Therefore, the dimensions of the rectangle that will maximize the enclosed
area are 75 meters by 150 meters.
24
Question 28
Question
A rectangular garden has a fixed perimeter of 80 meters. If the area of the
garden is to be maximized, determine the dimensions of the garden.
Solution
Step 1: Let’s denote the length of the garden as xmeters and the width as y
meters.
Step 2: We are given that the perimeter of the garden is fixed at 80 meters.
Therefore, we have the equation:
2x+ 2y= 80
Step 3: Simplifying the equation, we get:
x+y= 40
Step 4: We are also given that the area of the garden is to be maximized.
The area of a rectangle is given by A=xy.
Step 5: We can express yin terms of xfrom the equation x+y= 40 as
y= 40 −x.
Step 6: Substitute y= 40 −xinto the area formula A=xy to get:
A=x(40 −x)
Step 7: Simplifying the area equation, we have:
A= 40x−x2
Step 8: To maximize the area, we need to find the critical points. Taking
the derivative of the area function:
dA
dx = 40 −2x
Step 9: Setting the derivative to zero to find the critical points:
40 −2x= 0
2x= 40
x= 20
Step 10: Substituting x= 20 back into the equation x+y= 40 to find y:
20 + y= 40
y= 20
Step 11: Therefore, the dimensions of the garden that maximize the area
are 20 meters by 20 meters.
25
Question 29
Question
A rectangular garden has a length that is 4 meters longer than its width. If the
area of the garden is 72 square meters, what are the dimensions of the garden?
Solution
Let xrepresent the width of the garden in meters. Therefore, the length of the
garden is x+ 4 meters.
Step 1: The area of a rectangle is given by the formula A= length ×width.
We are given that the area of the garden is 72 square meters. Therefore, we
have the equation:
x(x+ 4) = 72
Step 2: Simplifying the equation:
x2+ 4x= 72
x2+ 4x−72 = 0
Step 3: We can solve this quadratic equation by factoring:
x2+ 9x−8x−72 = 0
x(x+ 9) −8(x+ 9) = 0
(x−8)(x+ 9) = 0
Step 4: Setting each factor to zero gives us two possible values for the width
of the garden:
x−8 = 0 =⇒x= 8 or x+ 9 = 0 =⇒x=−9
Step 5: Since the width cannot be negative, the width of the garden is 8
meters. The length is then:
x+ 4 = 8 + 4 = 12 meters
Therefore, the dimensions of the garden are 8 meters by 12 meters.
Question 30
Question
A rectangular garden is three times as long as it is wide. If the perimeter of the
garden is 48 meters, find the dimensions of the garden.
26
Solution
Step 1: Let’s denote the width of the garden as wmeters. Since the length is
three times the width, the length of the garden is 3wmeters.
Step 2: The perimeter of a rectangle is given by the formula P= 2(l+w),
where lis the length and wis the width. Substituting the values we have,
48 = 2(3w+w)
Step 3: Simplifying the equation, we get
48 = 2(4w)
48 = 8w
Step 4: Solving for w, we find
w= 6
Step 5: Now, we can find the length of the garden by substituting w= 6
back into l= 3w:
l= 3(6) = 18
Step 6: Therefore, the dimensions of the garden are: width = 6 meters and
length = 18 meters.
Question 31
Question
A rectangular garden has a length 5 meters longer than its width. If the area
of the garden is 144 square meters, find the dimensions of the garden.
Solution
Let the width of the garden be xmeters. Then, the length of the garden is x+5
meters.
Step 1: Write an equation for the area of the garden in terms of x:
x(x+ 5) = 144
Step 2: Expand the left side of the equation:
x2+ 5x= 144
Step 3: Rearrange the equation to form a quadratic equation in standard
form:
x2+ 5x−144 = 0
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Step 4: Solve the quadratic equation using the quadratic formula: The
quadratic formula is given by x=−b±√b2−4ac
2a, where a= 1, b= 5, and
c=−144.
x=−5±p52−4(1)(−144)
2(1)
x=−5±√25 + 576
2
x=−5±√601
2
x=−5±√601
2
Step 5: Based on the problem, we should choose the positive value for x
since the dimensions must be positive. Therefore,
x=−5 + √601
2≈8.89
Step 6: Calculate the length by adding 5 to x:
x+ 5 ≈8.89 + 5 = 13.89
So, the dimensions of the garden are approximately 8.89 meters by 13.89
meters.
Question 32
Question
A rectangular garden is 20 meters longer than it is wide. If the area of the
garden is 525 square meters, find the dimensions of the garden.
Solution
Step 1: Let’s denote the width of the garden as xmeters. Since the garden is 20
meters longer than it is wide, the length of the garden would be x+ 20 meters.
Step 2: The area of a rectangle is given by the formula A= length ×width.
Substituting the given values into this formula, we have:
x(x+ 20) = 525
Step 3: Expanding the left side of the equation, we get:
x2+ 20x= 525
Step 4: Rearranging this equation into standard quadratic form, we have:
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x2+ 20x−525 = 0
Step 5: To solve this quadratic equation, we can factorize it. The factors of
−525 that add up to 20 are 35 and −15. Therefore, we can rewrite the equation
as:
(x+ 35)(x−15) = 0
Step 6: Setting each factor to zero gives us two possible values for the width:
x+ 35 = 0 =⇒x=−35 (extraneous solution)
x−15 = 0 =⇒x= 15
Step 7: Since the width cannot be negative, the width of the garden is x= 15
meters.
Step 8: Substituting x= 15 back into the expression for the length (x+ 20)
gives us the length of the garden:
15 + 20 = 35
Step 9: Therefore, the dimensions of the garden are 15 meters by 35 meters.
Question 33
Question
A rectangular garden is twice as long as it is wide. If the area of the garden is
192 square meters, find the dimensions of the garden.
Solution
Step 1: Let’s denote the width of the garden as wmeters. Since the length is
twice the width, we can represent the length as 2wmeters.
Step 2: The area of a rectangle is given by the formula A= length ×width.
Substituting the given values into the formula, we have:
192 = 2w×w
Step 3: Simplifying the equation, we get:
192 = 2w2
Step 4: Divide both sides of the equation by 2 to solve for w2:
w2= 96
Step 5: Taking the square root of both sides, we find:
w=√96 = 4√6
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Step 6: Therefore, the width of the garden is 4√6 meters. Since the length
is twice the width, the length is:
2w= 2 ×4√6=8√6
Step 7: Thus, the dimensions of the garden are width = 4√6 meters and
length = 8√6 meters.
Question 34
Question
A rectangular garden with an area of 2400 square meters is to be surrounded
by a walkway of uniform width. If the walking area is to be 10 meters wide,
find the dimensions of the entire garden.
Solution
Let the width of the garden be xmeters. Since the area of the garden is given
by the product of its length and width, we have:
Step 1: Setting up the equation to represent the area of the garden:
x(x+ 20) = 2400
Step 2: Expand the left side of the equation:
x2+ 20x= 2400
Step 3: Rearrange the equation into a standard quadratic form:
x2+ 20x−2400 = 0
Step 4: Factor the quadratic equation:
(x+ 60)(x−40) = 0
Step 5: Solve for the possible values of xby setting each factor to zero:
x+ 60 = 0 =⇒x=−60 or x−40 = 0 =⇒x= 40
Step 6: Since the width cannot be negative, the width of the garden is 40
meters. Now, we can find the length of the garden:
Length = x+ 20 = 40 + 20 = 60
Therefore, the dimensions of the entire garden are 40 meters by 60 meters.
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Question 35
Question
A rectangular garden is 20 meters long and 15 meters wide. A path of uniform
width is to be built around the garden. If the total area of the garden and the
path is 600 square meters, find the width of the path.
Solution
Let the width of the path be xmeters.
Step 1: The total area of the garden and the path is the sum of the area of
the garden and the area of the path. Set up an equation based on this fact.
Total area = Area of garden + Area of path
Step 2: Calculate the area of the garden.
Area of garden = length ×width = 20 ×15 square meters
Area of garden = 300 square meters
Step 3: Calculate the dimensions of the entire region including the path.
The length of the entire region is 20 + 2xmeters (added width on both sides)
and the width is 15 + 2xmeters. Thus, the area of the entire region including
the path is
Area of region = (20 + 2x)(15 + 2x) square meters
Step 4: Form an equation using the total area given.
600 = 300 + (20 + 2x)(15 + 2x)
Step 5: Solve the equation by expanding and simplifying.
600 = 300 + 300 + 50x+ 40x+ 4x2
4x2+ 90x−300 = 0
Step 6: Factor the quadratic equation.
4x2+ 90x−300 = 0
x2+ 22.5x−75 = 0
(x+ 25)(x−3) = 0
Step 7: Find the possible solutions for x.
x=−25 (extraneous solution) or x= 3
Step 8: Check the solution. As the width cannot be negative, the width of
the path is 3 meters.
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