MATH 110 - INTERMEDIATE ALGEBRA Applications of quadratic equations Question Bank Set 2

MATH 110 - INTERMEDIATE

ALGEBRA - Applications of quadratic

equations

Question Bank - Set 2

Liberty University

Question 1

Question

A farmer wants to create a rectangular pen for his animals using one side of his

barn as one side of the pen, and 60 meters of fencing for the other three sides.

If the area of the pen is 600 square meters, what are the dimensions of the pen?

Solution

Let xbe the width of the pen and ybe the length of the pen.

Step 1: Write down the equations based on the given information. Since one

side of the pen is against the barn, we only need to consider the fencing for the

other three sides. The perimeter of the rectangular pen would be 2x+y= 60,

and the area would be xy = 600.

Step 2: Solve the first equation for yin terms of x. From the perimeter

equation, we can solve for yas y= 60 −2x.

Step 3: Substitute the expression for yinto the area equation. Substitute

y= 60 −2xinto the area equation to get x(60 −2x) = 600.

Step 4: Simplify and solve the quadratic equation. Expanding the left side

gives 60x−2x2= 600, which simplifies to −2x2+ 60x−600 = 0.

Step 5: Find the solutions for x. Divide the equation by −2 to simplify it

to x2−30x+ 300 = 0. We can factor this equation as (x−20)(x−10) = 0.

Step 6: Determine the dimensions of the pen. Since x= 20 implies y=

60−2(20) = 20, and x= 10 implies y= 60−2(10) = 40, the possible dimensions

are 20m by 20m or 10m by 40m.

Question 2

Question

A rectangular garden is to be created using 120 feet of fencing. If one side of

the garden is along a river and does not need fencing, find the dimensions of

the garden that will maximize the area.

Solution

Step 1: Let’s denote the length of the two parallel sides of the rectangular

garden as xand the width of the garden as y. Step 2: Since one side of the

garden is along the river and does not need fencing, the total amount of fencing

used is x+ 2y= 120. Step 3: Solve the equation x+ 2y= 120 for xto get

x= 120 −2y. Step 4: The area of the garden, A, is given by A=xy. Step 5:

Substitute x= 120 −2yinto the area formula to get A= (120 −2y)y. Step 6:

Expand the expression for the area to get A= 120y−2y2. Step 7: To find the

dimensions that maximize the area, we need to find the critical points. Take the

derivative of Awith respect to yand set it equal to 0. Step 8: dA

dy = 120 −4y.

Step 9: Set dA

dy = 0 and solve for y: 120 −4y= 0 ⇒y= 30. Step 10: Find the

corresponding value of x:x= 120 −2(30) = 60. Step 11: So, the dimensions of

the garden that maximize the area are 60 feet by 30 feet.

Question 3

Question

A company produces and sells a certain product. The revenue Rgenerated

by selling xunits of the product can be modeled by the quadratic equation

R(x) = −2x2+ 50x. Find the number of units that must be sold to maximize

the revenue.

Solution

Step 1: To find the number of units that must be sold to maximize the revenue,

we need to find the vertex of the quadratic function R(x) = −2x2+ 50x.

Step 2: The x-coordinate of the vertex of a quadratic function in the form

ax2+bx +cis given by x=−b

2a.

Step 3: In this case, a=−2 and b= 50. So, the x-coordinate of the vertex

is x=−50

2(−2) =50

4= 12.5.

Step 4: Therefore, the number of units that must be sold to maximize the

revenue is 12.5 units. However, since we can’t sell a fraction of a unit, we need

to round to the nearest whole number.

Step 5: So, the company should sell 13 units of the product to maximize the

revenue.

Question 4

Question

A farmer plans to create a rectangular vegetable garden by adding length to one

side of his existing square garden. He needs 180 meters of fencing material to

enclose both gardens. If the length of the rectangular garden is twice the length

of the square garden, find the dimensions of both gardens.

Solution

Let xbe the length of one side of the square garden. Since the length of the

rectangular garden is twice the length of the square garden, the length of the

rectangular garden is 2x. Since the square garden has four equal sides, the total

length of fencing material used for the square garden is 4x. The total length

of fencing material used for the rectangular garden is 2x+ 2(2x) = 6x. Given

that the total length of fencing material used for both gardens is 180 meters,

we have the equation:

4x+ 6x= 180

10x= 180

x= 18

So, the length of one side of the square garden is 18 meters, and the length of

the rectangular garden is 2(18) = 36 meters.

Therefore, the dimensions of the square garden are 18 ×18 meters, and the

dimensions of the rectangular garden are 36 ×18 meters.

Question 5

Question

A rectangular garden is to be created along the side of a house. If 40 meters

of fencing are available, what dimensions should the garden have in order to

maximize the area?

Solution

Let xrepresent the width of the garden and yrepresent the length of the garden.

Step 1: Write the perimeter equation. The perimeter of the rectangular

garden is given by:

2x+y= 40

Step 2: Write the area equation. The area of the rectangular garden is

given by:

A=xy

Step 3: Express yin terms of x. From the perimeter equation, y= 40 −2x.

Substitute this into the area equation:

A=x(40 −2x) = 40x−2x2

Step 4: Find the critical points. To find the maximum area, we need to

find the critical point of the function. Take the derivative of the area function:

dx = 40 −4x

Set this derivative equal to zero and solve for x:

40 −4x= 0

4x= 40

x= 10

Step 5: Determine the dimensions. Substitute x= 10 back into the perime-

ter equation to find y:

2(10) + y= 40

20 + y= 40

y= 20

Therefore, the dimensions that would maximize the area are 10 meters by

20 meters.

Question 6

Question

A rectangular garden is 12 meters longer than it is wide. If the area of the

garden is 180 square meters, find the dimensions of the garden.

Solution

Let wbe the width of the garden. Since the garden is 12 meters longer than it

is wide, the length of the garden is w+ 12.

We know that the area of a rectangle is given by the formula A= length ×

width. Given that the area of the garden is 180 square meters, we can set up

the following equation:

w(w+ 12) = 180

Step 1: Expand the left side of the equation.

w2+ 12w= 180

Step 2: Rearrange the quadratic equation into the standard form, w2+

12w−180 = 0.

Step 3: To solve the quadratic equation, we can either factor it or use

the quadratic formula. Let’s use the quadratic formula, which states that for

an equation in the form ax2+bx +c= 0, the solutions are given by x=

−b±√b2−4ac

2a.

In this case, a= 1, b= 12, and c=−180. Substituting into the quadratic

formula:

w=−12 ±p122−4(1)(−180)

2(1)

w=−12 ±√144 + 720

w=−12 ±√864

w=−12 ±24

The possible values for ware w1= 6 and w2=−30. Since the width cannot

be negative, we take w= 6.

Step 4: Find the length of the garden. The length of the garden is w+12 =

6 + 12 = 18 meters.

Step 5: State the dimensions of the garden. Therefore, the dimensions of

the garden are width 6 meters and length 18 meters.

Question 7

Question

A rectangular garden has a fixed perimeter of 40 meters. The length of the

garden is 6 meters longer than the width. Find the dimensions of the garden

that maximize the area.

Solution

Let’s denote the width of the garden as wmeters. Since the length is 6 meters

longer than the width, we can express the length in terms of was l=w+ 6.

Step 1: Write an equation based on the fixed perimeter of the garden. The

perimeter Pof a rectangle is given by 2×(length + width). Since the perimeter

is fixed at 40 meters, we have:

2(w+ (w+ 6)) = 40

Step 2: Find the value of w. Simplify the equation:

2w+ 2w+ 12 = 40

4w+ 12 = 40

4w= 28

w= 7

Step 3: Find the length of the garden. Using the value of w, we can find

the length:

l=w+ 6 = 7 + 6 = 13

Step 4: Calculate the area of the garden as a function of w. The area Aof

a rectangle is given by length ×width. Substituting w= 7 into the expression

for A, we have:

A(w) = 7 ×13 −91

Step 5: Determine the maximum area. To find the dimensions that maxi-

mize the area, we need to find the maximum of the quadratic function A(w).

The maximum occurs at the vertex of the parabola, which is given by w=

−b/(2a), where the quadratic function is in the form y=ax2+bx +c. In this

case, our quadratic function is A(w) = −w2+ 20w−91, so a=−1, b= 20.

w=−20

2(−1) = 10

Step 6: Find the maximum area. Now that we have the value of wthat

maximizes the area, we can find the corresponding length and calculate the

maximum area:

l= 10 + 6 = 16

Amax = 10 ×16 = 160

Therefore, the dimensions of the garden that maximize the area are 10 meters

by 16 meters, with a maximum area of 160 square meters.

Question 8

Question

A rectangular garden has a length that is 4 meters more than its width. The

area of the garden is 144 square meters. Find the dimensions of the garden.

Solution

Let’s denote the width of the rectangular garden as wmeters. Since the length

is 4 meters more than the width, we can express the length as w+ 4 meters.

Step 1: Write the equation for the area of the garden. The area of a rectangle

is given by the formula: Area = length ×width. Substitute the given values

into the formula:

144 = (w+ 4)w

Step 2: Simplify the equation. Expand the right side of the equation:

144 = w2+ 4w

Rearrange the equation into a quadratic equation form by setting it equal to

zero:

w2+ 4w−144 = 0

Step 3: Solve the quadratic equation. We can solve this quadratic equation

by factoring or using the quadratic formula. Let’s use the quadratic formula:

w=−b±√b2−4ac

where a= 1, b= 4, and c=−144.

w=−4±p42−4(1)(−144)

2(1)

w=−4±√16 + 576

w=−4±√592

w=−4±24.33

Step 4: Determine the possible values for the width. Since the width cannot

be negative, we take the positive value:

w=−4 + 24.33

2=20.33

2= 10.17

Step 5: Calculate the length. The length of the rectangular garden is w+ 4

meters.

Length = 10.17 + 4 = 14.17

Therefore, the dimensions of the rectangular garden are width = 10.17 me-

ters and length = 14.17 meters.

Question 9

Question

A ball is thrown horizontally from the top of a building 40 m high with an initial

speed of 20 m/s. The ball lands on the ground tseconds later. Find the total

time the ball is in the air.

Solution

Let’s denote the total time the ball is in the air as Tseconds.

Step 1: Find the time it takes for the ball to reach the ground horizontally.

This can be done by using the fact that the ball was thrown horizontally. The

equation to find the time using the displaced distance xis given by:

x=vt

Where xis the horizontal distance, vis the initial horizontal velocity, and tis

the time taken. In this case, x= 40 m and v= 20 m/s. Therefore, we can

rearrange the formula to solve for t.

t=x

v=40

20 = 2 seconds

Step 2: Find the total time the ball is in the air. Since the ball was in

the air for 2 seconds horizontally, we need to consider the time it was also in

the air vertically. This can be found using the height of the building and the

acceleration due to gravity. The equation to find the time using the height his

given by:

h=1

2gt2

Where his the height, gis the acceleration due to gravity, and tis the total

time taken. In this case, h= 40 m and g= 9.8 m/s2. We will solve for t.

40 = 1

2×9.8×T2

T2=40 ×2

9.8

T2=80

9.8

T2≈8.1633

T≈√8.1633 ≈2.8596

Therefore, the total time the ball is in the air is approximately 2.8596 sec-

onds.

Question 10

Question

A rectangular garden is to be enclosed with 120 meters of fencing. If one side

of the garden is against the side of a barn and so doesn’t need fencing, find the

dimensions of the garden that will maximize the area.

Solution

Let xrepresent the width of the rectangular garden and yrepresent the length

of the rectangular garden. Since one side of the garden is against the barn, only

three sides need to be fenced with a total length of 120 meters. Therefore, the

perimeter of the garden is given by the equation x+ 2y= 120.

To find the dimensions that will maximize the area, we need to express the

area of the garden in terms of one variable. The area Aof the garden is given

by A=xy.

Step 1: Express yin terms of xusing the perimeter equation. Substitute

y=120−x

2into the area equation:

A(x) = x120 −x

2

Step 2: Expand A(x) and simplify. Expand the area function:

A(x) = 1

2(120x−x2)

A(x) = 60x−1

2x2

Step 3: Find the critical points of A(x). To find the critical points, set the

derivative of A(x) equal to zero:

A′(x) = 60 −x= 0

x= 60

Step 4: Determine the dimensions that maximize the area. Since x= 60,

we can find yusing the perimeter equation:

y=120 −60

2= 30

Therefore, the dimensions of the garden that will maximize the area are width

60 meters and length 30 meters.

Question 11

Question

A rectangular field is 60 meters longer than it is wide. If the length of the field

is increased by 10 meters and the width is increased by 5 meters, the new area

is 1920 square meters. Find the dimensions of the original rectangular field.

Solution

Let the width of the original field be wmeters. Then, the length of the original

field is w+ 60 meters.

Step 1: Represent the area of the original field in terms of w. The area of

a rectangle is given by the formula A= length ×width. For the original field,

the area is given by:

Aoriginal = (w+ 60)w=w2+ 60w

Step 2: Represent the area of the new field in terms of w. After increasing

the length by 10 meters and the width by 5 meters, the new dimensions are

w+ 60 + 10 = w+ 70 meters for the length and w+ 5 meters for the width.

Therefore, the area of the new field is given by:

Anew = (w+ 70)(w+ 5) = w2+ 75w+ 350

Step 3: Equate the areas and solve for w. Since the new area is 1920 square

meters, we have the equation:

w2+ 75w+ 350 = 1920

w2+ 75w−1570 = 0

Step 4: Solve the quadratic equation to find the width of the original field.

Using the quadratic formula w=−b±√b2−4ac

2awith a= 1, b= 75, and c=

−1570, we get:

w=−75 ±p752−4(−1570)

w=−75 ±√5625 + 6280

w=−75 ±√11905

Step 5: Since negative width does not make sense in this context, the width

of the original field is:

w=−75 + √11905

2≈25.96 meters

Step 6: Calculate the length of the original field. Using the width w≈

25.96, the length of the original field is:

Length = w+ 60 ≈25.96 + 60 = 85.96 meters

Step 7: The dimensions of the original rectangular field are approximately

25.96 meters by 85.96 meters.

Question 12

Question

A person standing on a bridge 100 meters above the ground sees a boat floating

directly underneath. The person drops a stone from the bridge just as the boat

passes below. The stone hits the water 5 seconds later. Find the speed of the

boat.

Solution

Let’s denote the speed of the boat as vm/s.

Step 1: Determine the time taken for the stone to hit the water. The

distance the stone falls is equal to the height of the bridge, which is 100 meters.

Using the formula for distance fallen under gravity s=1

2gt2, where sis the

distance fallen, gis the acceleration due to gravity (approximately 9.8 m/s2),

and tis the time taken, we have:

100 = 1

2×9.8×t2

Solving for tgives us t=q200

9.8≈4.52 seconds.

Step 2: Find the distance covered by the boat during the time it took the

stone to fall. Given the time taken for the stone to hit the water is 5 seconds,

the boat would have traveled a distance of 5vmeters during this time.

Step 3: Set up an equation using the relative positions of the boat and

the stone. Since the boat was directly below the person when the stone was

dropped, the boat’s distance traveled in 5 seconds would be the same as the

distance the stone fell. Hence, we have the equation: 5v= 100

Step 4: Solve the equation to find the speed of the boat. From the equation

5v= 100, we can solve for v:

v=100

5= 20 m/s

Therefore, the speed of the boat is 20 m/s.

Question 13

Question

A rectangular garden is to be constructed along a riverbank. If 480 meters of

fencing are available and the riverbank will serve as one side of the garden, what

dimensions will maximize the area of the garden?

Solution

Let xbe the width of the garden perpendicular to the riverbank, and ybe the

length of the garden along the riverbank. The perimeter of the garden is given

as 480 meters, so we have the equation:

2x+y= 480

The area of the garden is given by A=xy. We need to find the dimensions

that will maximize this area.

Step 1: Solve the perimeter equation for yin terms of x.

2x+y= 480

y= 480 −2x

Step 2: Rewrite the area equation in terms of xonly.

A=x(480 −2x)

A= 480x−2x2

Step 3: Find the critical points of the area function by taking the derivative

with respect to xand setting it equal to zero.

dx = 480 −4x

0 = 480 −4x

4x= 480

x= 120

Step 4: Determine if this critical point gives a maximum area. To do this,

we can use the second derivative test.

d2A

dx2=−4

Since the second derivative is negative, the critical point at x= 120 gives a

maximum area.

Step 5: Find the dimensions that maximize the area. Substitute x= 120

back into the equation y= 480 −2x.

y= 480 −2(120) = 240

So, the dimensions that maximize the area of the garden are 120 meters by

240 meters.

Question 14

Question

A rectangular garden is to be created using a wall as one side and a hedge for

the other three sides. The area of the garden is 120 square meters. If the length

of the garden is xmeters and the cost of the hedge is 3 dollars per meter, express

the cost of the hedge as a function of x.

Solution

Let the length of the garden be xmeters and the width be ymeters. Since the

area of the garden is given by xy = 120, we can express yin terms of x.

Step 1: Express the width yin terms of the length x. Given that xy = 120,

solve for y:

y=120

Step 2: Determine the perimeter of the garden. The perimeter of the

rectangular garden is given by P= 2(x+y+x) = 2(2x+120

x) = 4x+240

Step 3: Express the cost of the hedge as a function of the length x. Since

the cost of the hedge is 3 dollars per meter, the cost function C(x) is given by

C(x) = 3 ×perimeter = 3(4x+240

x).

Therefore, the cost of the hedge as a function of the length xis C(x) =

12x+ 720/x.

Question 15

Question

A rectangular garden is to be constructed against the wall of a house. The

owner has 60 meters of fencing material to enclose three sides of the garden,

with the house acting as the fourth side. If the length of the garden is 3 times

its width, what should the dimensions of the garden be to maximize the area?

Solution

Let’s denote the width of the garden as wmeters. Since the length of the garden

is 3 times its width, the length can be represented as 3wmeters. We are looking

to maximize the area of the rectangular garden.

Step 1: Express the total length of fencing in terms of the width w. The

total length of fencing is the sum of the lengths of the three sides, two sides of

length weach and one side of length 3w. This can be expressed as:

2w+ 3w= 60

Step 2: Solve for the width w. Combining like terms, we get:

5w= 60

w=60

w= 12

Step 3: Find the dimensions and area of the garden. Since the length is 3

times the width, the length is:

3w= 3(12) = 36 meters

Therefore, the dimensions of the garden should be 12 meters by 36 meters to

maximize the area.

Step 4: Calculate the maximum area. The area Aof the rectangle can be

calculated as:

A= width ×length

Substitute the values of width and length:

A= 12 ×36

A= 432 square meters

Thus, the maximum area of the rectangular garden is 432 square meters

when the dimensions are 12 meters by 36 meters.

Question 16

Question

A rectangular garden has a fixed perimeter of 60 meters. If the length of the

garden is 6 more than its width, find the dimensions of the garden that maximize

its area.

Solution

Step 1: Let wbe the width of the garden. Since the length is 6 more than the

width, the length can be represented as w+ 6.

Step 2: The perimeter of the garden is given by 2w+ 2(w+ 6) = 60, which

simplifies to 4w+ 12 = 60. Solving for w, we find w= 12.

Step 3: Substituting w= 12 back into the expression for the length, we can

find the length: w+ 6 = 12 + 6 = 18.

Step 4: The area of the garden can be found by multiplying the width and

length, A=w×(w+ 6) = 12 ×18 = 216 square meters.

Step 5: To find the dimensions that maximize the area, we take the derivative

of the area function with respect to wand set it equal to 0.

Step 6: The derivative of the area function, A= 12w+ 6w, is A′= 12 + 6 =

18.

Step 7: Setting A′= 18 equal to 0 and solving for wgives us w= 0. However,

a width of 0 is not feasible in this context.

Step 8: Since the area of a rectangle is maximized when its width is half its

perimeter, and the width is 12 meters, the dimensions that maximize the area

are 12 meters by 18 meters.

Question 17

Question

A rectangular garden has a fixed perimeter of 60 meters. If the length of the

garden is 5 meters longer than the width, what dimensions should the garden

have to maximize the area it encloses?

Solution

Let xbe the width of the garden. Since the length is 5 meters longer than the

width, the length can be represented as x+ 5.

Step 1: Write an equation for the perimeter of the garden in terms of x. The

perimeter of a rectangle is given by the formula 2(length) + 2(width). Since the

length is x+ 5 and the width is x, the perimeter can be expressed as:

2(x+ 5) + 2x= 60

Step 2: Simplify the equation and solve for x.

2x+ 10 + 2x= 60

4x+ 10 = 60

4x= 50

x= 12.5

Step 3: Calculate the length. Since the length is x+ 5, the length of the

garden is 12.5 + 5 = 17.5 meters.

Step 4: Calculate the area of the garden. The area of a rectangle is given

by the formula length ×width. In this case, the area is:

12.5×17.5 = 218.75 m2

Therefore, to maximize the area the garden encloses, the dimensions should

be 12.5 meters by 17.5 meters.

Question 18

Question

A rectangular garden is three times as long as it is wide. The garden is to be

expanded so that each side is increased by 5 meters. If the area of the new

garden is 189 square meters more than the original garden, find the dimensions

of the original garden.

Solution

Let wbe the original width of the garden and 3wbe the original length of the

garden.

Step 1: Express the areas of the original and new gardens in terms of the

dimensions.

The area of the original garden is A=w×3w= 3w2square meters.

The expanded garden will have dimensions (w+ 5) and (3w+ 5), so its area

will be (w+ 5)(3w+ 5) square meters.

Step 2: Write an equation based on the given information.

The area of the new garden is 189 square meters more than the original

garden, so we have the equation:

(w+ 5)(3w+ 5) = 3w2+ 189

Step 3: Solve the equation.

Expanding the left side of the equation gives:

3w2+ 5w+ 15w+ 25 = 3w2+ 189

3w2+ 20w+ 25 = 3w2+ 189

20w+ 25 = 189

20w= 164

w=164

w= 8.2

Step 4: Find the dimensions of the original garden.

Since the original length is three times the width, the original length is

3×8.2 = 24.6 meters.

Therefore, the original garden has dimensions 8.2 m ×24.6 m .

Question 19

Question

A rectangular garden is to be created with a fixed area of 1000 square meters.

One side of the garden will be bounded by a building, so only three sides need

to be enclosed by fencing. If the fencing costs

10 per meter for the two lengths

and

20 per meter for the width, find the dimensions of the garden that will

minimize the cost of the fencing.

Solution

Let xbe the length of the building side, ybe the width of the garden, and A

be the area of the garden.

Step 1: Write the equations for the area and cost of the fencing. The area

of the garden is given by A=xy = 1000 square meters. The cost Cof the

fencing is given by C= 10(2x) + 20y= 20x+ 20y.

Step 2: Express the width yin terms of the length xusing the area con-

straint. From the area constraint xy = 1000, we have y=1000

Step 3: Rewrite the cost equation in terms of one variable using the width

expression. Substitute the width expression into the cost equation to get C=

20x+ 20 1000

x.

Step 4: Find the critical points of the cost function. To find the critical

points, we differentiate the cost function with respect to xand set it to zero.

dx = 20 −20000

x2= 0

20x2−20000 = 0

x2= 1000

x=±√1000 = ±10√10

Step 5: Determine the values of xthat minimize the cost. Since the length

must be positive, we take x= 10√10. Calculate the corresponding width using

y=1000

10√10 = 10√10

Step 6: State the dimensions that minimize the cost. Therefore, the dimen-

sions of the garden that will minimize the cost of the fencing are length 10√10

meters and width 10√10 meters.

Question 20

Question

A rectangular garden is to be enclosed with a fence on three sides and a straight

stone wall on the fourth side. If the total length of the fence used is 40 meters,

find the dimensions of the garden that maximize the enclosed area.

Solution

Step 1: Let’s denote the length of the garden as xmeters and the width of the

garden as ymeters.

Step 2: The total length of the fence used is equal to the perimeter of the

garden, which is given by 40 = 2x+y.

Step 3: To maximize the enclosed area, we need to express the area of the

garden in terms of one variable. The area Aof the garden is given by A=xy.

Step 4: Express yin terms of xfrom the perimeter equation: y= 40 −2x.

Step 5: Substitute y= 40 −2xinto the area equation to get A=x(40 −2x).

Step 6: Simplify the expression: A= 40x−2x2.

Step 7: To find the dimensions that maximize the area, we need to find the

critical points of the area function. We take the derivative of Awith respect to

x:dA

dx = 40 −4x.

Step 8: Set the derivative equal to zero and solve for x: 40−4x= 0 ⇒x= 10.

Step 9: Check that this critical point is a maximum by using the second

derivative test. The second derivative of Ais d2A

dx2=−4 which is negative,

indicating a maximum.

Step 10: Calculate the corresponding value of y:y= 40 −2(10) = 20.

Step 11: Therefore, the dimensions of the garden that maximize the enclosed

area are length x= 10 meters and width y= 20 meters.

Question 21

Question

A rectangular garden is to be enclosed with a fence on three sides and a wall on

the fourth side. If the length of the garden is xmeters, and the total area en-

closed is 300 square meters, find the dimensions of the garden that will minimize

the amount of fencing used.

Solution

Let the width of the garden be ymeters. Since the total area enclosed is 300

square meters, we have the equation:

x∗y= 300

We want to minimize the amount of fencing used, which consists of three

sides (the length x, the width y, and the width again). Thus, the total amount

of fencing is given by the function:

F(x, y) = x+ 2y

To minimize F(x, y), we need to express Fin terms of a single variable.

Using the area equation, we can write the width yin terms of x:

y=300

Substitute this expression for yin the function F(x, y) to get F(x):

F(x) = x+ 2 ∗300

To minimize F(x), we find its critical points by setting its derivative to zero:

F′(x)=1−600

x2= 0

x2= 600

x=√600 = 10√6 m

To verify this critical point corresponds to a minimum, we can use the second

derivative test:

F′′(x) = 1200

Since F′′(x)>0, the critical point x= 10√6 corresponds to a minimum for the

function F(x).

Thus, the dimensions of the garden that minimize the amount of fencing

used are x= 10√6 meters (length) and y=300

x=300

10√6= 5√6 meters (width).

Question 22

Question

A rectangular garden is to be created using a brick wall on one side and a hedge

on the remaining three sides. If the area of the garden is 200 square meters and

the cost of the brick wall is

30 per meter while the cost of the hedge is

10 per

meter, find the dimensions of the garden that minimize the cost of enclosing it.

Solution

Let the length of the garden be xmeters and the width be ymeters.

Step 1: Write down the equations representing the information given.

The area of the garden is 200 square meters:

xy = 200

The cost of the brick wall is

30 per meter and the cost of the hedge is

per meter. The cost function C(x, y) for enclosing the garden is:

C(x, y) = 30x+ 10(2x+ 2y) = 30x+ 20x+ 20y= 50x+ 20y

Step 2: Express the cost function in terms of a single variable using the

area constraint.

From the area constraint xy = 200, we can express yin terms of x:y=200

Substitute this into the cost function:

C(x) = 50x+ 20 200

x= 50x+4000

Step 3: Find the critical points by differentiating C(x) with respect to x

and setting it equal to zero:

dx = 50 −4000

x2= 0

50 = 4000

x2=4000

x2= 80

x=√80 = 4√5

Step 4: Determine the corresponding yvalues using the area constraint:

y=200

x=200

4√5= 50 ·1

√5= 10√5

Step 5: Calculate the minimum cost:

C(4√5) = 50(4√5) + 20(10√5) = 200√5 + 200√5 = 400√5≈894.43

Therefore, the dimensions of the garden that minimize the cost of enclosing

it are length 4√5 meters and width 10√5 meters, with a minimum cost of

approximately

894.43.

Question 23

Question

A rectangular garden has one side 5 meters longer than the other. If the area

of the garden is 156 square meters, find the dimensions of the garden.

Solution

Let xbe the length of the shorter side of the garden in meters. Therefore, the

length of the longer side can be expressed as x+ 5.

Given that the area of the garden is 156 square meters, we have the equation:

x(x+ 5) = 156

We now solve this quadratic equation to find the dimensions of the garden.

Step 1: Expand the left side of the equation.

x2+ 5x= 156

Step 2: Rearrange the equation into standard form.

x2+ 5x−156 = 0

Step 3: Factor the quadratic equation.

(x+ 13)(x−8) = 0

Step 4: Solve for xby setting each factor to zero.

x+ 13 = 0 or x−8=0

Step 5: Solve for xin each equation. For x+ 13 = 0:

x=−13

Since the length of a side cannot be negative, we disregard this solution.

For x−8 = 0:

x= 8

Step 6: Calculate the length of the longer side.

Longer side = x+ 5 = 8 + 5 = 13

Step 7: Final Answer: The dimensions of the garden are 8 meters by 13

meters.

Question 24

Question

A rectangular garden is to be created using 100 meters of fencing. If one side of

the garden is along a straight river, find the dimensions of the garden that will

give the largest possible area.

Solution

Step 1: Let xbe the length of the garden perpendicular to the river, and let y

be the width of the garden along the river. We want to maximize the area of

the garden, which is given by A=x·y.

Step 2: The perimeter of the garden is given by 100 = 2x+y, since we have

two sides of length xand one side of length y(along the river).

Step 3: Solving the perimeter equation for y, we get y= 100 −2x.

Step 4: Substitute the expression for yinto the area equation: A=x(100 −

2x) = 100x−2x2.

Step 5: To find the maximum area, we need to find the critical points of A

by taking the derivative of Awith respect to xand setting it equal to zero:

dx = 100 −4x

Setting this equal to zero and solving for x, we get x= 25.

Step 6: To confirm that this point gives a maximum area, we will use the

second derivative test. d2A

dx2=−4

Since the second derivative is negative, x= 25 corresponds to a local maximum.

Step 7: Find the corresponding yvalue using y= 100 −2x:

y= 100 −2(25) = 50

Step 8: Therefore, the dimensions of the garden that will give the largest

possible area are 25 meters by 50 meters.

Question 25

Question

A rectangular garden is to be created with one side along a river and the other

sides enclosed with a fence. If only 200 meters of fencing is available and the

area of the garden needs to be 1500 square meters, find the dimensions of the

garden that maximize the area.

Solution

Let the length of the garden along the river be xmeters and the width be y

meters.

Step 1: Write the equations for the perimeter and the area of the garden.

The total length of fencing available is twice the length of the garden and the

width, with the length along the river requiring no fencing. Therefore, we can

write:

2y+x= 200

The area of the garden is given by:

xy = 1500

Step 2: Rewrite the perimeter equation in terms of y. Solve the perimeter

equation for xto get:

x= 200 −2y

Step 3: Substitute xfrom the previous step into the formula for the area.

Substitute x= 200 −2yinto the area equation to get:

(200 −2y)y= 1500

200y−2y2= 1500

Step 4: Rearrange the equation to form a quadratic equation. Rearrange

the equation to form a quadratic equation in standard form:

2y2−200y+ 1500 = 0

Step 5: Solve the quadratic equation to find the width y. Using the

quadratic formula, we find:

y=−(−200) ±p(−200)2−4(2)(1500)

2(2)

y=200 ±20√10

y= 50 ±5√10

Step 6: Determine the dimensions that maximize the area. Since yrepre-

sents the width of the garden, we choose the positive value for y:

y= 50 + 5√10

x= 200 −2(50 + 5√10)

x= 100 −10√10

Therefore, the dimensions that maximize the area of the garden are: Length

x= 100 −10√10 meters, Width y= 50 + 5√10 meters.

Question 26

Question

A rectangular garden is 20 meters long and 15 meters wide. A walkway of

uniform width is added around the garden, increasing the total area to 500

square meters. Find the width of the walkway.

Solution

Let xbe the width of the walkway.

Step 1: First, we find the total area of the original garden without the

walkway. This can be calculated as the product of the original length and

width:

Original Area = Length ×Width = 20 ×15 = 300 sq meters

Step 2: We can express the dimensions of the entire garden, including the

walkway, in terms of x: Length: 20 + 2xmeters Width: 15 + 2xmeters

Step 3: The total area of the garden with the walkway is given as 500

square meters. We can set up an equation using the total area formula:

Total Area = Length ×Width

500 = (20 + 2x)(15 + 2x)

Step 4: Expand the expression on the right side and simplify the quadratic

equation:

500 = 300 + 70x+ 40x+ 4x2

4x2+ 110x−200 = 0

Step 5: Solve the quadratic equation using the quadratic formula:

x=−110 ±p1102−4×4×(−200)

2×4

x=−110 ±√12100 + 3200

x=−110 ±√15300

x=−110 ±90

Step 6: The width of the walkway must be positive, so we take the positive

solution:

x=−110 + 90

8=−20

8=−2.5

Step 7: Since the width of the walkway cannot be negative, we must discard

the negative solution. Hence, the width of the walkway is 2.5 meters.

Question 27

Question

A rectangular garden is 30 meters long and 20 meters wide. The garden is

expanded by planting a uniform border of flowers around it, increasing the area

of the garden by 50

Solution

Let the width of the border be xmeters. The new dimensions of the garden

will be (30 + 2x) meters by (20 + 2x) meters.

Step 1: The original area of the garden is 30 ×20 = 600 square meters.

Step 2: The new area of the garden after adding the border is 600 +

0.5(600) = 900 square meters.

Step 3: The new area of the garden can also be expressed as the product

of its new dimensions:

(30 + 2x)(20 + 2x) = 900

Step 4: Expand the left side of the equation:

600 + 100x+ 60x+ 4x2= 900

Step 5: Simplify the equation:

4x2+ 160x−300 = 0

Step 6: To solve the quadratic equation 4x2+ 160x−300 = 0, we can

simplify it by dividing each term by 4:

x2+ 40x−75 = 0

Step 7: Use the quadratic formula to solve for x:

x=−b±√b2−4ac

where a= 1, b= 40, and c=−75.

Step 8: Substitute a,b, and cinto the formula and simplify to find the

solutions for x. The width of the border should be a positive number, so the

negative solution can be discarded.

Step 9: Calculate the width of the border to get the final answer.

Question 28

Question

A rectangular garden is 10 meters longer than it is wide. If the area of the

garden is 192 square meters, find the dimensions of the garden.

Solution

Let xbe the width of the garden (in meters). Since the garden is 10 meters

longer than it is wide, the length of the garden can be expressed as x+ 10.

We are given that the area of the garden is 192 square meters. The area of

a rectangle is given by the formula:

Area = length ×width

Substituting in the expressions for length and width:

x(x+ 10) = 192

Now, we will solve the quadratic equation to find the dimensions of the

garden.

Step 1: Expand the left side of the equation.

x2+ 10x= 192

Step 2: Rearrange the equation into standard form.

x2+ 10x−192 = 0

Step 3: Factor the quadratic equation.

(x−12)(x+ 16) = 0

Setting each factor to zero:

x−12 = 0 =⇒x= 12

x+ 16 = 0 =⇒x=−16

Since the width cannot be negative, the width of the garden is 12 meters.

Step 4: Find the length of the garden.

Length = x+ 10 = 12 + 10 = 22 meters.

Step 5: Write the final answer.

The dimensions of the garden are 12 meters by 22 meters.

Question 29

Question

A rectangular garden is to be constructed using a wall as one side and a hedge

as the other three sides. The length of the wall is 20 meters. The area of the

garden is to be 240 square meters. Find the dimensions of the garden that will

minimize the amount of hedge used.

Solution

Let xbe the length of the garden perpendicular to the wall, and let ybe the

width of the garden parallel to the wall. The area of the garden can be expressed

as A=xy = 240.

Step 1: Express the amount of hedge used in terms of xand y. The amount

of hedge used will be the sum of the lengths of the three sides that need hedge:

Hedge used = x+ 2y

Step 2: Express the amount of hedge used in terms of one variable. Since

A= 240, we can express yin terms of x:

y=240

Substitute yin the expression for the amount of hedge used:

Hedge used = x+ 2 240

x

Simplify the expression:

Hedge used = x+480

Step 3: Find the minimum amount of hedge used. To find the minimum

amount of hedge used, we can find the critical points of the function by finding

where the derivative is equal to zero:

dx(Hedge used) = 1 −480

x2= 0

Solve for x:

x2= 480

x=√480 = 4√30

Step 4: Determine the corresponding value of y. Using the relationship

y=240

x, we can find:

y=240

4√30 = 60√30

Step 5: Verify the minimum. To verify that this point is a minimum, we can

use the second derivative test. The second derivative of the function is positive,

confirming that the point is a minimum.

Step 6: State the dimensions of the garden that minimize the amount of

hedge used. Therefore, the dimensions of the garden that minimize the amount

of hedge used are 4√30 m ×60√30 m.

Question 30

Question

A farmer wants to build a rectangular enclosure for his animals using a long

barn as one side of the enclosure. If he has 100 meters of fencing material and

wants to maximize the area of the enclosure, what should the dimensions of the

enclosure be?

Solution

Let the length of the barn be xmeters, and the width of the enclosure be y

meters. Since the barn is used as one side, there are only three sides that need

fencing. Thus, the total amount of fencing material used is x+ 2y= 100.

To maximize the area of the enclosure, we need to maximize A=xy. From

the information given, we can express xin terms of yas x= 100 −2y. Substi-

tuting into the area formula gives A= (100 −2y)y= 100y−2y2.

To maximize the area, we need to find the critical points by taking the

derivative of Awith respect to yand setting it equal to zero.

dy = 100 −4y= 0

4y= 100

y= 25

To ensure that this value yields a maximum area, we need to check the

endpoints of the possible range for y: 0 ≤y≤50. At y= 0, x= 100 and A= 0,

and at y= 50, x= 0 and A= 0. So, the maximum area occurs when y= 25.

Therefore, the dimensions of the enclosure that maximize the area are 25

meters by 50 meters.

Question 31

Question

A rectangular garden has a fixed perimeter of 40 meters. The area of the garden

is to be maximized. Let xand yrepresent the dimensions of the garden. Write

an equation that represents the area of the garden in terms of xand y, and then

use quadratic equations to find the dimensions that maximize the area.

Solution

Step 1: Let’s write an equation for the area of the garden in terms of xand

y. The perimeter of the rectangular garden is given by 2x+ 2y= 40, which

simplifies to x+y= 20. Solving for y, we get y= 20 −x. The area of the

rectangular garden is given by A=xy. Substituting y= 20 −xinto this

equation gives A=x(20 −x).

Step 2: Expand the equation for the area. A= 20x−x2

Step 3: To find the dimensions that maximize the area, we need to find the

critical points of A. The critical points occur where the derivative of Ais zero.

dx = 20 −2x

Setting dA

dx = 0 and solving for x: 20 −2x= 0

2x= 20

x= 10

Step 4: To determine if x= 10 corresponds to a maximum or minimum,

we will use the second derivative test. d2A

dx2=−2 Since the second derivative is

negative, x= 10 corresponds to a maximum.

Step 5: Substitute x= 10 into y= 20 −xto find the corresponding value of

y.y= 20 −10

y= 10

Therefore, the dimensions that maximize the area of the garden are x= 10

meters and y= 10 meters.

Question 32

Question

A rectangular garden with a length of 20 meters and a width of 10 meters is

going to be expanded by the same amount on all sides to form a new rectangular

garden with an area of 400 square meters. Find the width of the expanded

garden.

Solution

Let the amount by which the garden is expanded on all sides be xmeters. Then

the new length of the garden will be 20 + 2xmeters and the new width will be

10 + 2xmeters.

The area of the expanded garden is the product of the new length and the

new width, which is given to be 400 square meters. Therefore, we have the

equation:

(20 + 2x)(10 + 2x) = 400

Expanding the left side gives:

200 + 40x+ 40x+ 4x2= 400

4x2+ 80x+ 200 = 400

4x2+ 80x−200 = 0

Now we need to solve this quadratic equation. We can simplify it by dividing

through by 4:

x2+ 20x−50 = 0

Next, we solve for xusing the quadratic formula x=−b±√b2−4ac

2a.

Step 1: Identify a,b, and cin the equation ax2+bx +c= 0.

Here, a= 1, b= 20, and c=−50.

Step 2: Substitute a= 1, b= 20, and c=−50 into the quadratic formula.

x=−20 ±p202−4∗1∗(−50)

2∗1

x=−20 ±√400 + 200

x=−20 ±√600

x=−20 ±10√6

Therefore, the width of the expanded garden is 10 + x= 10 + −20±10√6

Thus, the width could be 5 −5√6 or 15 + 5√6 meters.

Question 33

Question

A rectangular photograph is to be enlarged. The length of the photograph is to

be increased by 20

Solution

Let the original length of the photograph be Linches and the original width be

Winches.

Step 1: Calculate the area of the original photograph. The area of the

original photograph is given by Aoriginal =L×W.

Step 2: Calculate the new dimensions of the photograph. After the en-

largement, the new length will be 1.2L(increased by 20

Step 3: Write an equation for the area of the new photograph. The area of

the new photograph is given by Anew = 1.2L×0.8W= 0.96LW .

Step 4: Write down the equation given in the problem. We are given that

the area of the new photograph is 144 square inches larger than the original

photograph: 0.96LW =LW + 144.

Step 5: Solve the equation to find the dimensions of the original photograph.

Subtract LW from both sides of the equation: 0.96LW −LW = 144

0.04LW = 144

LW =144

0.04

LW = 3600

Thus, the area of the original photograph is 3600 square inches.

Step 6: Find the original dimensions. Since Aoriginal =L×W= 3600, and

we know from Step 1 that Aoriginal =LW , we have: LW = 3600.

This means the dimensions of the original photograph are 60 inches by 60

inches.

Question 34

Question

A rectangular garden is to be created along the side of a house. The area of the

garden must be 300 square meters, and one of the sides will be against the wall of

the house. If fencing costs 10permeter, whatdimensionsshouldthegardenhaveinordertominimizethecostofthef encing?

Solution

Let xbe the width of the garden (perpendicular to the house) and ybe the

length of the garden (parallel to the house). We are given that the area of the

garden is 300 square meters, so we have the equation xy = 300.

The cost of fencing is given by C= 10(2x+y), as there are two sides of

length xand one side of length y. We want to minimize this cost, i.e., find the

minimum of C.

Step 1: Express the cost function in terms of a single variable. From the

area equation xy = 300, we can express yin terms of xas y=300

x. Substitute

this into the cost function:

C= 10(2x+300

x) = 20x+3000

Step 2: Find the critical points by taking the derivative of Cwith respect

to x.

C′(x) = 20 −3000

Set C′(x) = 0 to find critical points:

20 −3000

x2= 0

3000

x2= 20

x2= 150

x=√150

Step 3: Determine if the critical point is a minimum. To determine if

x=√150 is a minimum, we need to check the second derivative.

C′′(x) = 6000

At x=√150, we have C′′(√150) = 6000

(√150)3=6000

150√150 =40

√150 >0, which

means x=√150 is a minimum.

Step 4: Calculate the corresponding length and minimum cost. From xy =

300, when x=√150:

y=300

√150 = 2√150

Therefore, the dimensions of the garden that minimize the cost of fencing

are √150 meters by 2√150 meters, with a minimum cost of 20√150 (or approx-

imately 120.0)m.

Question 35

Question

A rectangular garden is to be created along the side of a house. The area of

the garden must be 200 square meters. If the house wall will serve as one of the

sides of the garden, find the dimensions of the garden that would require the

least amount of fencing.