MATH 332 - Applications of derivatives Question
Bank
Question 1
A company has determined that the weekly demand function for their prod-
uct can be modeled by the equation:
p= 300 −0.5q
where pis the price per unit (in dollars) and qis the quantity of units sold
per week.
Task: Determine the rate at which revenue is changing with respect to the
number of units sold when 200 units are sold per week.
Solution:
Step 1: Establish the revenue function. Revenue Ris calculated as the
product of price pand quantity q:
R(q) = p×q
Using the demand function, the revenue function can be represented as:
R(q) = (300 −0.5q)×q
R(q) = 300q−0.5q2
Step 2: Differentiate the revenue function with respect to q. To find the
rate of change of revenue with respect to the number of units sold, take the
derivative of R(q) with respect to q:
dR
dq =d
dq (300q−0.5q2)
dR
dq = 300 −q
Step 3: Evaluate the derivative at q= 200. Substitute q= 200 into the
derivative to find the rate of change of revenue at that level of sales:
dR
dq
q=200
= 300 −200
1
dR
dq
q=200
= 100
Conclusion: The rate at which revenue is changing with respect to the num-
ber of units sold when 200 units are sold per week is
$
100 per unit. This means
that, at 200 units sold, for each additional unit sold, the revenue increases by
100.Question1
A company has determined that the weekly demand function for their prod-
uct can be modeled by the equation:
p= 300 −0.5q
where pis the price per unit (in dollars) and qis the quantity of units sold
per week.
Task: Determine the rate at which revenue is changing with respect to the
number of units sold when 200 units are sold per week.
Solution:
Step 1: Establish the revenue function. Revenue Ris calculated as the
product of price pand quantity q:
R(q) = p×q
Using the demand function, the revenue function can be represented as:
R(q) = (300 −0.5q)×q
R(q) = 300q−0.5q2
Step 2: Differentiate the revenue function with respect to q. To find the
rate of change of revenue with respect to the number of units sold, take the
derivative of R(q) with respect to q:
dR
dq =d
dq (300q−0.5q2)
dR
dq = 300 −q
Step 3: Evaluate the derivative at q= 200. Substitute q= 200 into the
derivative to find the rate of change of revenue at that level of sales:
dR
dq
q=200
= 300 −200
dR
dq
q=200
= 100
Conclusion: The rate at which revenue is changing with respect to the num-
ber of units sold when 200 units are sold per week is
$
100 per unit. This means
that, at 200 units sold, for each additional unit sold, the revenue increases by
100.
2
Question 2
Problem Statement: A company produces and sells handmade candles. The
price at which the candles can be sold is influenced by the number of candles
produced. Each candle is sold at a price given by the equation P(x) = 20 −0.1x
dollars, where xrepresents the number of candles produced and sold. The cost
to produce xcandles is given by C(x)=2x+ 100 dollars. Find the number of
candles xthat must be produced and sold to maximize the profit.
Solution Steps:
Step 1: Write down the equation for profit. Profit, Π(x), is given by revenue
minus cost. Revenue R(x) from selling xcandles at the price P(x) is R(x) =
xP (x).
Π(x) = R(x)−C(x) = x(20 −0.1x)−(2x+ 100)
Step 2: Simplify the profit function.
Π(x) = 20x−0.1x2−2x−100
Π(x) = −0.1x2+ 18x−100
Step 3: Find the derivative of Π(x) and set it to zero to find critical points.
Π′(x) = −0.2x+ 18
Set Π′(x) = 0 to find the critical points.
−0.2x+ 18 = 0
0.2x= 18
x= 90
Step 4: Use the second derivative test to confirm that this critical point
corresponds to a maximum.
Π′′ (x) = −0.2
Since Π′′ (x)<0, this indicates a local maximum.
Step 5: Conclude the optimal number of candles to produce and sell. The
company maximizes profit when it produces and sells 90 candles.
Step 6: Calculate the maximum profit. Substitute x= 90 back into the
profit function:
Π(90) = −0.1(90)2+ 18(90) −100
Π(90) = −0.1(8100) + 1620 −100
Π(90) = −810 + 1620 −100
Π(90) = 710
Thus, the maximum profit is
$
710.
3
Conclusion To maximize profit, the company should produce and sell 90
candles, resulting in a maximum profit of
$
710. Question 2: Maximum
Profit Problem
Problem Statement: A company produces and sells handmade can-
dles. The price at which the candles can be sold is influenced by the
number of candles produced. Each candle is sold at a price given by
the equation P(x) = 20 −0.1xdollars, where xrepresents the number
of candles produced and sold. The cost to produce xcandles is given
by C(x)=2x+ 100 dollars. Find the number of candles xthat must
be produced and sold to maximize the profit.
Solution Steps:
Step 1: Write down the equation for profit. Profit, Π(x), is given
by revenue minus cost. Revenue R(x)from selling xcandles at the
price P(x)is R(x) = xP (x).
Π(x) = R(x)−C(x) = x(20 −0.1x)−(2x+ 100)
Step 2: Simplify the profit function.
Π(x) = 20x−0.1x2−2x−100
Π(x) = −0.1x2+ 18x−100
Step 3: Find the derivative of Π(x)and set it to zero to find critical
points.
Π′(x) = −0.2x+ 18
Set Π′(x)=0to find the critical points.
−0.2x+ 18 = 0
0.2x= 18
x= 90
Step 4: Use the second derivative test to confirm that this critical
point corresponds to a maximum.
Π′′ (x) = −0.2
Since Π′′ (x)<0, this indicates a local maximum.
Step 5: Conclude the optimal number of candles to produce and
sell. The company maximizes profit when it produces and sells 90
candles.
Step 6: Calculate the maximum profit. Substitute x= 90 back into
the profit function:
Π(90) = −0.1(90)2+ 18(90) −100
Π(90) = −0.1(8100) + 1620 −100
4
Π(90) = −810 + 1620 −100
Π(90) = 710
Thus, the maximum profit is
$
710.
Conclusion To maximize profit, the company should produce and
sell 90 candles, resulting in a maximum profit of
$
710.
Question 3
Problem Statement:
A company manufactures and sells a product at a price that decreases with
the quantity sold. The price function is given by p(x) = 600 −0.4xdollars per
unit, where xis the number of units sold. The cost to manufacture xunits is
represented by the function C(x) = 12000 + 100xdollars.
Determine the number of units, x, the company should sell to maximize its
profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Write the profit function. Profit, P(x), is the revenue minus the
cost. The revenue, R(x), from selling xunits at p(x) dollars per unit is given
by R(x) = x·p(x). Therefore:
R(x) = x(600 −0.4x) = 600x−0.4x2
P(x) = R(x)−C(x) = 600x−0.4x2−(12000 + 100x) = 500x−0.4x2−12000
Step 2: Calculate the derivative of the profit function.
P′(x) = 500 −0.8x
Step 3: Set the derivative equal to zero to find the critical points.
500 −0.8x= 0
0.8x= 500
x=500
0.8= 625
Step 4: Use the second derivative test to determine if this critical point is a
maximum. Find P′′ (x):
P′′ (x) = −0.8
Since P′′ (x)<0, the function is concave down at x= 625, indicating a local
maximum.
Step 5: Calculate the maximum profit. Substitute x= 625 into P(x):
P(625) = 500(625) −0.4(625)2−12000
= 312500 −0.4×390625 −12000
5
= 312500 −156250 −12000
= 144250
Step 6: Answer the question. The company should sell 625 units to maximize
its profit. The maximum profit is 144,250.Question3 : MaximizingP rof it
Problem Statement:
A company manufactures and sells a product at a price that decreases with
the quantity sold. The price function is given by p(x) = 600 −0.4xdollars per
unit, where xis the number of units sold. The cost to manufacture xunits is
represented by the function C(x) = 12000 + 100xdollars.
Determine the number of units, x, the company should sell to maximize its
profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Write the profit function. Profit, P(x), is the revenue minus the
cost. The revenue, R(x), from selling xunits at p(x) dollars per unit is given
by R(x) = x·p(x). Therefore:
R(x) = x(600 −0.4x) = 600x−0.4x2
P(x) = R(x)−C(x) = 600x−0.4x2−(12000 + 100x) = 500x−0.4x2−12000
Step 2: Calculate the derivative of the profit function.
P′(x) = 500 −0.8x
Step 3: Set the derivative equal to zero to find the critical points.
500 −0.8x= 0
0.8x= 500
x=500
0.8= 625
Step 4: Use the second derivative test to determine if this critical point is a
maximum. Find P′′ (x):
P′′ (x) = −0.8
Since P′′ (x)<0, the function is concave down at x= 625, indicating a local
maximum.
Step 5: Calculate the maximum profit. Substitute x= 625 into P(x):
P(625) = 500(625) −0.4(625)2−12000
= 312500 −0.4×390625 −12000
= 312500 −156250 −12000
= 144250
Step 6: Answer the question. The company should sell 625 units to maximize
its profit. The maximum profit is 144,250.
6
Question 4
Problem Statement:
A farmer has 2400 feet of fencing and wants to enclose a rectangular area
next to a river. There is no need to fence the side along the river. Calculate
the dimensions of the rectangle that maximize the enclosed area. What is the
maximum area?
Step-by-step Solution:
Step 1: Define the Variables
Let’s denote: - Las the length of the fence parallel to the river. - Was the
width of the fence perpendicular to the river.
Step 2: Write the Equation for Perimeter
Since there is no need to fence along the river, the required amount of fencing
will only concern two widths and one length:
2W+L= 2400 feet
Step 3: Express Lin Terms of W
From the equation above, solve for L:
L= 2400 −2W
Step 4: Write the Equation for Area
The area Aof the rectangle can be expressed as:
A=L×W
Substitute Lfrom step 3:
A= (2400 −2W)W
A= 2400W−2W2
Step 5: Differentiate the Area Function
To find the critical points, first, differentiate Awith respect to W:
dA
dW = 2400 −4W
Step 6: Set the Derivative Equal to Zero to Find Critical Points
Now, set dA
dW = 0 to find the critical points:
2400 −4W= 0
4W= 2400
W= 600 feet
Step 7: Determine LWhen W= 600 Feet
7
Substitute W= 600 into the equation from step 3:
L= 2400 −2×600
L= 1200 feet
Step 8: Confirm Maximization
Using either the second derivative test or the nature of the problem (a max-
imum area enclosed by a fixed amount of fencing suggests one maximum), we
can infer: d2A
dW 2=−4
Since d2A
dW 2is negative, we have a maximum point.
Step 9: Calculate the Maximum Area
Substitute back the values of Land Winto the area formula:
A=L×W= 1201 ×600
A= 720,000 square feet
Conclusion:
The dimensions that maximize the enclosed area are 1200 feet parallel to the
river and 600 feet perpendicular to the river, giving a maximum area of 720,000
square feet. Question 4: Applications of Derivatives - Optimization
Problem
Problem Statement:
A farmer has 2400 feet of fencing and wants to enclose a rectan-
gular area next to a river. There is no need to fence the side along
the river. Calculate the dimensions of the rectangle that maximize
the enclosed area. What is the maximum area?
Step-by-step Solution:
Step 1: Define the Variables
Let’s denote: - Las the length of the fence parallel to the river. -
Was the width of the fence perpendicular to the river.
Step 2: Write the Equation for Perimeter
Since there is no need to fence along the river, the required amount
of fencing will only concern two widths and one length:
2W+L= 2400 feet
Step 3: Express Lin Terms of W
From the equation above, solve for L:
L= 2400 −2W
Step 4: Write the Equation for Area
The area Aof the rectangle can be expressed as:
A=L×W
8
Substitute Lfrom step 3:
A= (2400 −2W)W
A= 2400W−2W2
Step 5: Differentiate the Area Function
To find the critical points, first, differentiate Awith respect to W:
dA
dW = 2400 −4W
Step 6: Set the Derivative Equal to Zero to Find Critical Points
Now, set dA
dW = 0 to find the critical points:
2400 −4W= 0
4W= 2400
W= 600 feet
Step 7: Determine LWhen W= 600 Feet
Substitute W= 600 into the equation from step 3:
L= 2400 −2×600
L= 1200 feet
Step 8: Confirm Maximization
Using either the second derivative test or the nature of the problem
(a maximum area enclosed by a fixed amount of fencing suggests one
maximum), we can infer:
d2A
dW 2=−4
Since d2A
dW 2is negative, we have a maximum point.
Step 9: Calculate the Maximum Area
Substitute back the values of Land Winto the area formula:
A=L×W= 1201 ×600
A= 720,000 square feet
Conclusion:
The dimensions that maximize the enclosed area are 1200 feet
parallel to the river and 600 feet perpendicular to the river, giving a
maximum area of 720,000 square feet.
9
Question 5
Problem: A company finds that the cost, in dollars, of producing xunits of
a particular product is given by the cost function C(x) = 100x+2000. The com-
pany sells these units at a price of 150each.Determinetheproductionlevelthatmaximizesthecompany′sprof it.
Step-by-Step Solution:
Step 1: Define the revenue and profit functions. - The revenue function R(x)
from selling xunits is given by the price per unit times the number of units sold,
i.e., R(x) = 150x. - The profit function P(x) is the revenue minus the cost, so
P(x) = R(x)−C(x) = 150x−(100x+ 2000).
Step 2: Simplify the profit function.
P(x) = 150x−100x−2000 = 50x−2000
Step 3: Find the derivative of the profit function.
P′(x) = d
dx (50x−2000) = 50
Step 4: Set the first derivative equal to zero to find critical points.
50 = 0
Since 50 = 0, there are no critical points, suggesting that the function does not
have a maximum or minimum value based on this derivative alone.
Step 5: Analyze endpoints and practical constraints. Since a negative pro-
duction level isn’t feasible, consider x≥0. The derivative tells us the function
is increasing over the interval, and therefore the profit function is continually
increasing as production increases.
Step 6: Consider practical limits to production levels. While mathematically
the profit increases indefinitely with production, practically, the company will
have a maximum production capacity M(say the company can produce at most
10,000 units due to machinery, labor, or materials constraints).
Step 7: Substitute practical maximum production level into profit function.
P(10000) = 50 ×10000 −2000 = 500000 −2000 = 498000
Thus, under practical constraints, the profit is maximized at the production
level of 10,000 units.
Conclusion: The company’s profit maximizes at the highest feasible produc-
tion level within practical constraints, assumed here as 10,000 units, yielding a
profit of 498,000.Question5 : Applicationsof Derivatives
Problem: A company finds that the cost, in dollars, of producing xunits of
a particular product is given by the cost function C(x) = 100x+2000. The com-
pany sells these units at a price of 150each.Determinetheproductionlevelthatmaximizesthecompany′sprof it.
Step-by-Step Solution:
Step 1: Define the revenue and profit functions. - The revenue function R(x)
from selling xunits is given by the price per unit times the number of units sold,
10
i.e., R(x) = 150x. - The profit function P(x) is the revenue minus the cost, so
P(x) = R(x)−C(x) = 150x−(100x+ 2000).
Step 2: Simplify the profit function.
P(x) = 150x−100x−2000 = 50x−2000
Step 3: Find the derivative of the profit function.
P′(x) = d
dx (50x−2000) = 50
Step 4: Set the first derivative equal to zero to find critical points.
50 = 0
Since 50 = 0, there are no critical points, suggesting that the function does not
have a maximum or minimum value based on this derivative alone.
Step 5: Analyze endpoints and practical constraints. Since a negative pro-
duction level isn’t feasible, consider x≥0. The derivative tells us the function
is increasing over the interval, and therefore the profit function is continually
increasing as production increases.
Step 6: Consider practical limits to production levels. While mathematically
the profit increases indefinitely with production, practically, the company will
have a maximum production capacity M(say the company can produce at most
10,000 units due to machinery, labor, or materials constraints).
Step 7: Substitute practical maximum production level into profit function.
P(10000) = 50 ×10000 −2000 = 500000 −2000 = 498000
Thus, under practical constraints, the profit is maximized at the production
level of 10,000 units.
Conclusion: The company’s profit maximizes at the highest feasible produc-
tion level within practical constraints, assumed here as 10,000 units, yielding a
profit of 498,000.
Question 6
Problem Statement: A company produces and sells handmade candles. The
weekly profit function Pin dollars, as a function of the number of candles x
sold, is given by:
P(x) = −0.1x2+ 40x−300
To maximize their profit, the company wants to determine the number of
candles they should produce and sell each week.
Questions:
a) Find the derivative of the profit function, P(x).
b) Determine the critical point(s) for the profit function.
11
c) Use the second derivative test to confirm whether the critical point is a
maximum.
d) What is the maximum weekly profit, and how many candles must be sold
to achieve this profit?
Solution:
a) Finding the derivative of the profit function, P(x):
Step-by-Step:
1. Given Function:
P(x) = −0.1x2+ 40x−300
2. Differentiate P(x) with respect to x:
P′(x) = d
dx (−0.1x2+ 40x−300)
P′(x) = −0.2x+ 40
b) Determining the critical point(s):
Step-by-Step:
1. Set the derivative P′(x) to zero:
−0.2x+ 40 = 0
2. Solve for x:
−0.2x=−40
x= 200
c) Using the second derivative test:
Step-by-Step:
1. Find the second derivative P′′ (x):
P′′ (x) = d
dx (−0.2x+ 40)
P′′ (x) = −0.2
2. Evaluate P′′ (x) at the critical point x= 200:
P′′ (200) = −0.2
Since P′′ (200) <0, the function P(x) has a local maximum at x= 200.
d) Maximum weekly profit and number of candles:
Step-by-Step:
1. Substitute x= 200 into P(x) to find maximum profit:
P(200) = −0.1(200)2+ 40(200) −300
P(200) = −0.1(40000) + 8000 −300
P(200) = −4000 + 8000 −300
12
P(200) = 3700
Thus, the maximum weekly profit is 3700, andtoachievethis, thecompanymustproduceandsell200candleseachweek.Question6 :
Applicationsof Derivatives −F indingM aximumP rofit
Problem Statement: A company produces and sells handmade candles. The
weekly profit function Pin dollars, as a function of the number of candles x
sold, is given by:
P(x) = −0.1x2+ 40x−300
To maximize their profit, the company wants to determine the number of
candles they should produce and sell each week.
Questions:
a) Find the derivative of the profit function, P(x).
b) Determine the critical point(s) for the profit function.
c) Use the second derivative test to confirm whether the critical point is a
maximum.
d) What is the maximum weekly profit, and how many candles must be sold
to achieve this profit?
Solution:
a) Finding the derivative of the profit function, P(x):
Step-by-Step:
1. Given Function:
P(x) = −0.1x2+ 40x−300
2. Differentiate P(x) with respect to x:
P′(x) = d
dx (−0.1x2+ 40x−300)
P′(x) = −0.2x+ 40
b) Determining the critical point(s):
Step-by-Step:
1. Set the derivative P′(x) to zero:
−0.2x+ 40 = 0
2. Solve for x:
−0.2x=−40
x= 200
c) Using the second derivative test:
Step-by-Step:
1. Find the second derivative P′′ (x):
P′′ (x) = d
dx (−0.2x+ 40)
13
P′′ (x) = −0.2
2. Evaluate P′′ (x) at the critical point x= 200:
P′′ (200) = −0.2
Since P′′ (200) <0, the function P(x) has a local maximum at x= 200.
d) Maximum weekly profit and number of candles:
Step-by-Step:
1. Substitute x= 200 into P(x) to find maximum profit:
P(200) = −0.1(200)2+ 40(200) −300
P(200) = −0.1(40000) + 8000 −300
P(200) = −4000 + 8000 −300
P(200) = 3700
Thus, the maximum weekly profit is 3700, andtoachievethis, thecompanymustproduceandsell200candleseachweek.
Question 7
Problem: A company manufactures a product that has a demand equation
given by p= 300−2q, where pis the price per unit in dollars and qis the quantity
sold per month. Find the rate at which revenue is changing with respect to price
when the price is 100.
Solution:
Step 1: Write the revenue function. The revenue Ris the product of the
price pand the quantity q. From the given demand equation p= 300 −2q, we
can solve for qin terms of p:
q=300 −p
2
The revenue function Rcan then be expressed as:
R=p·q=p·300 −p
2
R=300p−p2
2
Step 2: Differentiate the revenue function with respect to p. To find how
the revenue changes with respect to price, we differentiate Rwith respect to p:
dR
dp =d
dp 300p−p2
2
dR
dp =1
2(300 −2p)
14
dR
dp = 150 −p
Step 3: Evaluate dR
dp when p= 100. Substituting p= 100 into the derivative:
dR
dp = 150 −100
dR
dp = 50
Conclusion: The rate at which revenue is changing with respect to price
when the price is 100is50 per dollar increase in price. This means that if the
price per unit increases by 1, therevenuewillincreasebyapproximately50 when
the price is at 100.Question7 : Applicationsof Derivatives
Problem: A company manufactures a product that has a demand equation
given by p= 300−2q, where pis the price per unit in dollars and qis the quantity
sold per month. Find the rate at which revenue is changing with respect to price
when the price is 100.
Solution:
Step 1: Write the revenue function. The revenue Ris the product of the
price pand the quantity q. From the given demand equation p= 300 −2q, we
can solve for qin terms of p:
q=300 −p
2
The revenue function Rcan then be expressed as:
R=p·q=p·300 −p
2
R=300p−p2
2
Step 2: Differentiate the revenue function with respect to p. To find how
the revenue changes with respect to price, we differentiate Rwith respect to p:
dR
dp =d
dp 300p−p2
2
dR
dp =1
2(300 −2p)
dR
dp = 150 −p
Step 3: Evaluate dR
dp when p= 100. Substituting p= 100 into the derivative:
dR
dp = 150 −100
dR
dp = 50
15
Conclusion: The rate at which revenue is changing with respect to price
when the price is 100is50 per dollar increase in price. This means that if the
price per unit increases by 1, therevenuewillincreasebyapproximately50 when
the price is at 100.
Question 8
Problem Statement: A company produces and sells handmade candles. The
price per candle that consumers are willing to pay decreases as the quantity of
candles sold increases, according to the demand function:
p(x) = 20 −0.5x
where pis the price in dollars per candle and xis the number of candles sold
per day. The cost per day, C(x), of producing xcandles is given by:
C(x) = 100 + 3x
Find the number of candles the company should produce and sell each day
to maximize profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Determine the Revenue Function. Revenue, R, is the product of the
number of units sold, x, and the price per unit, p(x). So,
R(x) = x·p(x) = x(20 −0.5x)
R(x) = 20x−0.5x2
Step 2: Determine the Profit Function. Profit, Π, is the difference between
Revenue and Cost. So,
Π(x) = R(x)−C(x)
Π(x) = (20x−0.5x2)−(100 + 3x)
Π(x) = 20x−0.5x2−100 −3x
Π(x) = −0.5x2+ 17x−100
Step 3: Calculate the Derivative of the Profit Function. To find the profit
maximization point, take the derivative of Π(x) with respect to x, and set it to
zero.
Π′(x) = −x+ 17
Set Π′(x) = 0:
−x+ 17 = 0
x= 17
16
Step 4: Verify Maximum Profit Using Second Derivative Test. Compute the
second derivative of Π(x):
Π′′ (x) = −1
Since Π′′ (x) = −1 is negative, the function has a local maximum at x= 17.
Step 5: Calculate Maximum Profit. Plug x= 17 back into the profit function
to find the maximum profit.
Π(17) = −0.5(17)2+ 17(17) −100
Π(17) = −0.5×289 + 289 −100
Π(17) = −144.5 + 289 −100
Π(17) = 44.5
Conclusion: The company should produce and sell 17 candles per day to
maximize their profit, which will be 44.5perday.Question8 : Applicationsof Derivatives−
MaximumP rofitP roblem
Problem Statement: A company produces and sells handmade candles. The
price per candle that consumers are willing to pay decreases as the quantity of
candles sold increases, according to the demand function:
p(x) = 20 −0.5x
where pis the price in dollars per candle and xis the number of candles sold
per day. The cost per day, C(x), of producing xcandles is given by:
C(x) = 100 + 3x
Find the number of candles the company should produce and sell each day
to maximize profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Determine the Revenue Function. Revenue, R, is the product of the
number of units sold, x, and the price per unit, p(x). So,
R(x) = x·p(x) = x(20 −0.5x)
R(x) = 20x−0.5x2
Step 2: Determine the Profit Function. Profit, Π, is the difference between
Revenue and Cost. So,
Π(x) = R(x)−C(x)
Π(x) = (20x−0.5x2)−(100 + 3x)
Π(x) = 20x−0.5x2−100 −3x
Π(x) = −0.5x2+ 17x−100
17
Step 3: Calculate the Derivative of the Profit Function. To find the profit
maximization point, take the derivative of Π(x) with respect to x, and set it to
zero.
Π′(x) = −x+ 17
Set Π′(x) = 0:
−x+ 17 = 0
x= 17
Step 4: Verify Maximum Profit Using Second Derivative Test. Compute the
second derivative of Π(x):
Π′′ (x) = −1
Since Π′′ (x) = −1 is negative, the function has a local maximum at x= 17.
Step 5: Calculate Maximum Profit. Plug x= 17 back into the profit function
to find the maximum profit.
Π(17) = −0.5(17)2+ 17(17) −100
Π(17) = −0.5×289 + 289 −100
Π(17) = −144.5 + 289 −100
Π(17) = 44.5
Conclusion: The company should produce and sell 17 candles per day to
maximize their profit, which will be 44.5perday.
Question 9
A company produces and sells an item. The price per item, p, in dollars,
can be expressed as a function of the number of items, x, sold and is given by
the equation p(x) = 40 −0.1x. The revenue Ris the product of the number of
items sold and the price per item. The task is to find how many items should
be sold to maximize revenue.
Step-by-step Solution:
Step 1: Identify the revenue function Since revenue Ris the product of the
number of items sold, x, and the price per item, p(x), you first write down the
expression for Rusing the function for p(x):
R(x) = x×p(x) = x×(40 −0.1x)
R(x) = 40x−0.1x2
Step 2: Take the derivative of the revenue function Differentiate R(x) with
respect to xto find R′(x):
R′(x) = 40 −0.2x
18
Step 3: Set the first derivative equal to zero to find critical points Set R′(x) =
0 and solve for x:
40 −0.2x= 0
0.2x= 40
x= 200
Step 4: Evaluate the concavity using the second derivative or critical points
To ensure this value of xgives a maximum, check the sign of the second deriva-
tive, R′′ (x):
R′′ (x) = −0.2
Since R′′ (x)<0, the function is concave down at x= 200, indicating a local
maximum.
Step 5: Conclusion The company will maximize revenue when they sell 200
items. The maximum revenue can also be calculated:
R(200) = 40(200) −0.1(200)2= 8000 −4000 = 4000
Hence, the maximum revenue is 4000.
This provides a guided solution to maximizing the revenue for the company
based on the given price function. Question 9: Maximizing Revenue
A company produces and sells an item. The price per item, p, in
dollars, can be expressed as a function of the number of items, x, sold
and is given by the equation p(x) = 40 −0.1x. The revenue Ris the
product of the number of items sold and the price per item. The task
is to find how many items should be sold to maximize revenue.
Step-by-step Solution:
Step 1: Identify the revenue function Since revenue Ris the prod-
uct of the number of items sold, x, and the price per item, p(x), you
first write down the expression for Rusing the function for p(x):
R(x) = x×p(x) = x×(40 −0.1x)
R(x) = 40x−0.1x2
Step 2: Take the derivative of the revenue function Differentiate
R(x)with respect to xto find R′(x):
R′(x) = 40 −0.2x
Step 3: Set the first derivative equal to zero to find critical points
Set R′(x) = 0 and solve for x:
40 −0.2x= 0
0.2x= 40
x= 200
19
Step 4: Evaluate the concavity using the second derivative or crit-
ical points To ensure this value of xgives a maximum, check the sign
of the second derivative, R′′ (x):
R′′ (x) = −0.2
Since R′′ (x)<0, the function is concave down at x= 200, indicating a
local maximum.
Step 5: Conclusion The company will maximize revenue when they
sell 200 items. The maximum revenue can also be calculated:
R(200) = 40(200) −0.1(200)2= 8000 −4000 = 4000
Hence, the maximum revenue is 4000.
This provides a guided solution to maximizing the revenue for the
company based on the given price function.
Question 10
Problem Statement: A company produces and sells handmade bracelets.
The price per bracelet can be modeled by the demand function p(x) = 20−0.5x,
where pis the price in dollars and xis the number of bracelets sold per day.
The daily cost to produce xbracelets is given by C(x) = 40 + 2x. Determine
the number of bracelets the company should produce and sell to maximize its
daily revenue.
Solution:
Step 1: Define the Revenue Function The revenue function R(x) is defined as
the product of the number of units sold and the price per unit. Mathematically,
this is expressed as:
R(x) = x·p(x)
Substituting the demand function:
R(x) = x·(20 −0.5x) = 20x−0.5x2
Step 2: Compute the Derivative of the Revenue Function To find the maxi-
mum revenue, we need to find the critical points of R(x). This is done by taking
the derivative of R(x) and setting it to zero:
R′(x) = 20 −x
Set R′(x) = 0 to find the critical points:
20 −x= 0 =⇒x= 20
Step 3: Determine Whether This Critical Point Is a Maximum To confirm
whether this critical point x= 20 is a maximum, use the second derivative test.
Compute R′′ (x):
R′′ (x) = −1
20
Since R′′ (20) = −1 is negative, the function has a local maximum at x= 20.
Step 4: Evaluate the Maximum Revenue Plug x= 20 back into the revenue
function to find the maximum revenue:
R(20) = 20 ×20 −0.5×202= 400 −200 = 200
So, the maximum revenue is 200.
Step 5: Conclusion The company should produce and sell 20 bracelets daily
to maximize its revenue, which will be 200perdayatthislevelof production.
This answer outlines how the applications of derivatives, specifically using
the first and second derivative tests, are crucial in real-world business scenarios
like maximizing revenue. Question 10: Maximum Revenue Problem
Problem Statement: A company produces and sells handmade
bracelets. The price per bracelet can be modeled by the demand
function p(x) = 20 −0.5x, where pis the price in dollars and xis
the number of bracelets sold per day. The daily cost to produce x
bracelets is given by C(x) = 40+2x. Determine the number of bracelets
the company should produce and sell to maximize its daily revenue.
Solution:
Step 1: Define the Revenue Function The revenue function R(x)is
defined as the product of the number of units sold and the price per
unit. Mathematically, this is expressed as:
R(x) = x·p(x)
Substituting the demand function:
R(x) = x·(20 −0.5x) = 20x−0.5x2
Step 2: Compute the Derivative of the Revenue Function To find
the maximum revenue, we need to find the critical points of R(x).
This is done by taking the derivative of R(x)and setting it to zero:
R′(x) = 20 −x
Set R′(x) = 0 to find the critical points:
20 −x= 0 =⇒x= 20
Step 3: Determine Whether This Critical Point Is a Maximum
To confirm whether this critical point x= 20 is a maximum, use the
second derivative test. Compute R′′ (x):
R′′ (x) = −1
Since R′′ (20) = −1is negative, the function has a local maximum at
x= 20.
21
Step 4: Evaluate the Maximum Revenue Plug x= 20 back into the
revenue function to find the maximum revenue:
R(20) = 20 ×20 −0.5×202= 400 −200 = 200
So, the maximum revenue is 200.
Step 5: Conclusion The company should produce and sell 20 bracelets
daily to maximize its revenue, which will be 200perdayatthislevelof production.
This answer outlines how the applications of derivatives, specifi-
cally using the first and second derivative tests, are crucial in real-
world business scenarios like maximizing revenue.
22
Question 2
Problem Statement: A company produces and sells handmade candles. The
price at which the candles can be sold is influenced by the number of candles
produced. Each candle is sold at a price given by the equation P(x) = 20 −0.1x
dollars, where xrepresents the number of candles produced and sold. The cost
to produce xcandles is given by C(x)=2x+ 100 dollars. Find the number of
candles xthat must be produced and sold to maximize the profit.
Solution Steps:
Step 1: Write down the equation for profit. Profit, Π(x), is given by revenue
minus cost. Revenue R(x) from selling xcandles at the price P(x) is R(x) =
xP (x).
Π(x) = R(x)−C(x) = x(20 −0.1x)−(2x+ 100)
Step 2: Simplify the profit function.
Π(x) = 20x−0.1x2−2x−100
Π(x) = −0.1x2+ 18x−100
Step 3: Find the derivative of Π(x) and set it to zero to find critical points.
Π′(x) = −0.2x+ 18
Set Π′(x) = 0 to find the critical points.
−0.2x+ 18 = 0
0.2x= 18
x= 90
Step 4: Use the second derivative test to confirm that this critical point
corresponds to a maximum.
Π′′ (x) = −0.2
Since Π′′ (x)<0, this indicates a local maximum.
Step 5: Conclude the optimal number of candles to produce and sell. The
company maximizes profit when it produces and sells 90 candles.
Step 6: Calculate the maximum profit. Substitute x= 90 back into the
profit function:
Π(90) = −0.1(90)2+ 18(90) −100
Π(90) = −0.1(8100) + 1620 −100
Π(90) = −810 + 1620 −100
Π(90) = 710
Thus, the maximum profit is
$
710.
3
Conclusion To maximize profit, the company should produce and sell 90
candles, resulting in a maximum profit of
$
710. Question 2: Maximum
Profit Problem
Problem Statement: A company produces and sells handmade can-
dles. The price at which the candles can be sold is influenced by the
number of candles produced. Each candle is sold at a price given by
the equation P(x) = 20 −0.1xdollars, where xrepresents the number
of candles produced and sold. The cost to produce xcandles is given
by C(x)=2x+ 100 dollars. Find the number of candles xthat must
be produced and sold to maximize the profit.
Solution Steps:
Step 1: Write down the equation for profit. Profit, Π(x), is given
by revenue minus cost. Revenue R(x)from selling xcandles at the
price P(x)is R(x) = xP (x).
Π(x) = R(x)−C(x) = x(20 −0.1x)−(2x+ 100)
Step 2: Simplify the profit function.
Π(x) = 20x−0.1x2−2x−100
Π(x) = −0.1x2+ 18x−100
Step 3: Find the derivative of Π(x)and set it to zero to find critical
points.
Π′(x) = −0.2x+ 18
Set Π′(x)=0to find the critical points.
−0.2x+ 18 = 0
0.2x= 18
x= 90
Step 4: Use the second derivative test to confirm that this critical
point corresponds to a maximum.
Π′′ (x) = −0.2
Since Π′′ (x)<0, this indicates a local maximum.
Step 5: Conclude the optimal number of candles to produce and
sell. The company maximizes profit when it produces and sells 90
candles.
Step 6: Calculate the maximum profit. Substitute x= 90 back into
the profit function:
Π(90) = −0.1(90)2+ 18(90) −100
Π(90) = −0.1(8100) + 1620 −100
4
Π(90) = −810 + 1620 −100
Π(90) = 710
Thus, the maximum profit is
$
710.
Conclusion To maximize profit, the company should produce and
sell 90 candles, resulting in a maximum profit of
$
710.
Question 3
Problem Statement:
A company manufactures and sells a product at a price that decreases with
the quantity sold. The price function is given by p(x) = 600 −0.4xdollars per
unit, where xis the number of units sold. The cost to manufacture xunits is
represented by the function C(x) = 12000 + 100xdollars.
Determine the number of units, x, the company should sell to maximize its
profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Write the profit function. Profit, P(x), is the revenue minus the
cost. The revenue, R(x), from selling xunits at p(x) dollars per unit is given
by R(x) = x·p(x). Therefore:
R(x) = x(600 −0.4x) = 600x−0.4x2
P(x) = R(x)−C(x) = 600x−0.4x2−(12000 + 100x) = 500x−0.4x2−12000
Step 2: Calculate the derivative of the profit function.
P′(x) = 500 −0.8x
Step 3: Set the derivative equal to zero to find the critical points.
500 −0.8x= 0
0.8x= 500
x=500
0.8= 625
Step 4: Use the second derivative test to determine if this critical point is a
maximum. Find P′′ (x):
P′′ (x) = −0.8
Since P′′ (x)<0, the function is concave down at x= 625, indicating a local
maximum.
Step 5: Calculate the maximum profit. Substitute x= 625 into P(x):
P(625) = 500(625) −0.4(625)2−12000
= 312500 −0.4×390625 −12000
5
= 312500 −156250 −12000
= 144250
Step 6: Answer the question. The company should sell 625 units to maximize
its profit. The maximum profit is 144,250.Question3 : MaximizingP rof it
Problem Statement:
A company manufactures and sells a product at a price that decreases with
the quantity sold. The price function is given by p(x) = 600 −0.4xdollars per
unit, where xis the number of units sold. The cost to manufacture xunits is
represented by the function C(x) = 12000 + 100xdollars.
Determine the number of units, x, the company should sell to maximize its
profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Write the profit function. Profit, P(x), is the revenue minus the
cost. The revenue, R(x), from selling xunits at p(x) dollars per unit is given
by R(x) = x·p(x). Therefore:
R(x) = x(600 −0.4x) = 600x−0.4x2
P(x) = R(x)−C(x) = 600x−0.4x2−(12000 + 100x) = 500x−0.4x2−12000
Step 2: Calculate the derivative of the profit function.
P′(x) = 500 −0.8x
Step 3: Set the derivative equal to zero to find the critical points.
500 −0.8x= 0
0.8x= 500
x=500
0.8= 625
Step 4: Use the second derivative test to determine if this critical point is a
maximum. Find P′′ (x):
P′′ (x) = −0.8
Since P′′ (x)<0, the function is concave down at x= 625, indicating a local
maximum.
Step 5: Calculate the maximum profit. Substitute x= 625 into P(x):
P(625) = 500(625) −0.4(625)2−12000
= 312500 −0.4×390625 −12000
= 312500 −156250 −12000
= 144250
Step 6: Answer the question. The company should sell 625 units to maximize
its profit. The maximum profit is 144,250.
6
Question 4
Problem Statement:
A farmer has 2400 feet of fencing and wants to enclose a rectangular area
next to a river. There is no need to fence the side along the river. Calculate
the dimensions of the rectangle that maximize the enclosed area. What is the
maximum area?
Step-by-step Solution:
Step 1: Define the Variables
Let’s denote: - Las the length of the fence parallel to the river. - Was the
width of the fence perpendicular to the river.
Step 2: Write the Equation for Perimeter
Since there is no need to fence along the river, the required amount of fencing
will only concern two widths and one length:
2W+L= 2400 feet
Step 3: Express Lin Terms of W
From the equation above, solve for L:
L= 2400 −2W
Step 4: Write the Equation for Area
The area Aof the rectangle can be expressed as:
A=L×W
Substitute Lfrom step 3:
A= (2400 −2W)W
A= 2400W−2W2
Step 5: Differentiate the Area Function
To find the critical points, first, differentiate Awith respect to W:
dA
dW = 2400 −4W
Step 6: Set the Derivative Equal to Zero to Find Critical Points
Now, set dA
dW = 0 to find the critical points:
2400 −4W= 0
4W= 2400
W= 600 feet
Step 7: Determine LWhen W= 600 Feet
7
Substitute W= 600 into the equation from step 3:
L= 2400 −2×600
L= 1200 feet
Step 8: Confirm Maximization
Using either the second derivative test or the nature of the problem (a max-
imum area enclosed by a fixed amount of fencing suggests one maximum), we
can infer: d2A
dW 2=−4
Since d2A
dW 2is negative, we have a maximum point.
Step 9: Calculate the Maximum Area
Substitute back the values of Land Winto the area formula:
A=L×W= 1201 ×600
A= 720,000 square feet
Conclusion:
The dimensions that maximize the enclosed area are 1200 feet parallel to the
river and 600 feet perpendicular to the river, giving a maximum area of 720,000
square feet. Question 4: Applications of Derivatives - Optimization
Problem
Problem Statement:
A farmer has 2400 feet of fencing and wants to enclose a rectan-
gular area next to a river. There is no need to fence the side along
the river. Calculate the dimensions of the rectangle that maximize
the enclosed area. What is the maximum area?
Step-by-step Solution:
Step 1: Define the Variables
Let’s denote: - Las the length of the fence parallel to the river. -
Was the width of the fence perpendicular to the river.
Step 2: Write the Equation for Perimeter
Since there is no need to fence along the river, the required amount
of fencing will only concern two widths and one length:
2W+L= 2400 feet
Step 3: Express Lin Terms of W
From the equation above, solve for L:
L= 2400 −2W
Step 4: Write the Equation for Area
The area Aof the rectangle can be expressed as:
A=L×W
8
Substitute Lfrom step 3:
A= (2400 −2W)W
A= 2400W−2W2
Step 5: Differentiate the Area Function
To find the critical points, first, differentiate Awith respect to W:
dA
dW = 2400 −4W
Step 6: Set the Derivative Equal to Zero to Find Critical Points
Now, set dA
dW = 0 to find the critical points:
2400 −4W= 0
4W= 2400
W= 600 feet
Step 7: Determine LWhen W= 600 Feet
Substitute W= 600 into the equation from step 3:
L= 2400 −2×600
L= 1200 feet
Step 8: Confirm Maximization
Using either the second derivative test or the nature of the problem
(a maximum area enclosed by a fixed amount of fencing suggests one
maximum), we can infer:
d2A
dW 2=−4
Since d2A
dW 2is negative, we have a maximum point.
Step 9: Calculate the Maximum Area
Substitute back the values of Land Winto the area formula:
A=L×W= 1201 ×600
A= 720,000 square feet
Conclusion:
The dimensions that maximize the enclosed area are 1200 feet
parallel to the river and 600 feet perpendicular to the river, giving a
maximum area of 720,000 square feet.
9
Question 5
Problem: A company finds that the cost, in dollars, of producing xunits of
a particular product is given by the cost function C(x) = 100x+2000. The com-
pany sells these units at a price of 150each.Determinetheproductionlevelthatmaximizesthecompany′sprof it.
Step-by-Step Solution:
Step 1: Define the revenue and profit functions. - The revenue function R(x)
from selling xunits is given by the price per unit times the number of units sold,
i.e., R(x) = 150x. - The profit function P(x) is the revenue minus the cost, so
P(x) = R(x)−C(x) = 150x−(100x+ 2000).
Step 2: Simplify the profit function.
P(x) = 150x−100x−2000 = 50x−2000
Step 3: Find the derivative of the profit function.
P′(x) = d
dx (50x−2000) = 50
Step 4: Set the first derivative equal to zero to find critical points.
50 = 0
Since 50 = 0, there are no critical points, suggesting that the function does not
have a maximum or minimum value based on this derivative alone.
Step 5: Analyze endpoints and practical constraints. Since a negative pro-
duction level isn’t feasible, consider x≥0. The derivative tells us the function
is increasing over the interval, and therefore the profit function is continually
increasing as production increases.
Step 6: Consider practical limits to production levels. While mathematically
the profit increases indefinitely with production, practically, the company will
have a maximum production capacity M(say the company can produce at most
10,000 units due to machinery, labor, or materials constraints).
Step 7: Substitute practical maximum production level into profit function.
P(10000) = 50 ×10000 −2000 = 500000 −2000 = 498000
Thus, under practical constraints, the profit is maximized at the production
level of 10,000 units.
Conclusion: The company’s profit maximizes at the highest feasible produc-
tion level within practical constraints, assumed here as 10,000 units, yielding a
profit of 498,000.Question5 : Applicationsof Derivatives
Problem: A company finds that the cost, in dollars, of producing xunits of
a particular product is given by the cost function C(x) = 100x+2000. The com-
pany sells these units at a price of 150each.Determinetheproductionlevelthatmaximizesthecompany′sprof it.
Step-by-Step Solution:
Step 1: Define the revenue and profit functions. - The revenue function R(x)
from selling xunits is given by the price per unit times the number of units sold,
10
i.e., R(x) = 150x. - The profit function P(x) is the revenue minus the cost, so
P(x) = R(x)−C(x) = 150x−(100x+ 2000).
Step 2: Simplify the profit function.
P(x) = 150x−100x−2000 = 50x−2000
Step 3: Find the derivative of the profit function.
P′(x) = d
dx (50x−2000) = 50
Step 4: Set the first derivative equal to zero to find critical points.
50 = 0
Since 50 = 0, there are no critical points, suggesting that the function does not
have a maximum or minimum value based on this derivative alone.
Step 5: Analyze endpoints and practical constraints. Since a negative pro-
duction level isn’t feasible, consider x≥0. The derivative tells us the function
is increasing over the interval, and therefore the profit function is continually
increasing as production increases.
Step 6: Consider practical limits to production levels. While mathematically
the profit increases indefinitely with production, practically, the company will
have a maximum production capacity M(say the company can produce at most
10,000 units due to machinery, labor, or materials constraints).
Step 7: Substitute practical maximum production level into profit function.
P(10000) = 50 ×10000 −2000 = 500000 −2000 = 498000
Thus, under practical constraints, the profit is maximized at the production
level of 10,000 units.
Conclusion: The company’s profit maximizes at the highest feasible produc-
tion level within practical constraints, assumed here as 10,000 units, yielding a
profit of 498,000.
Question 6
Problem Statement: A company produces and sells handmade candles. The
weekly profit function Pin dollars, as a function of the number of candles x
sold, is given by:
P(x) = −0.1x2+ 40x−300
To maximize their profit, the company wants to determine the number of
candles they should produce and sell each week.
Questions:
a) Find the derivative of the profit function, P(x).
b) Determine the critical point(s) for the profit function.
11
c) Use the second derivative test to confirm whether the critical point is a
maximum.
d) What is the maximum weekly profit, and how many candles must be sold
to achieve this profit?
Solution:
a) Finding the derivative of the profit function, P(x):
Step-by-Step:
1. Given Function:
P(x) = −0.1x2+ 40x−300
2. Differentiate P(x) with respect to x:
P′(x) = d
dx (−0.1x2+ 40x−300)
P′(x) = −0.2x+ 40
b) Determining the critical point(s):
Step-by-Step:
1. Set the derivative P′(x) to zero:
−0.2x+ 40 = 0
2. Solve for x:
−0.2x=−40
x= 200
c) Using the second derivative test:
Step-by-Step:
1. Find the second derivative P′′ (x):
P′′ (x) = d
dx (−0.2x+ 40)
P′′ (x) = −0.2
2. Evaluate P′′ (x) at the critical point x= 200:
P′′ (200) = −0.2
Since P′′ (200) <0, the function P(x) has a local maximum at x= 200.
d) Maximum weekly profit and number of candles:
Step-by-Step:
1. Substitute x= 200 into P(x) to find maximum profit:
P(200) = −0.1(200)2+ 40(200) −300
P(200) = −0.1(40000) + 8000 −300
P(200) = −4000 + 8000 −300
12
P(200) = 3700
Thus, the maximum weekly profit is 3700, andtoachievethis, thecompanymustproduceandsell200candleseachweek.Question6 :
Applicationsof Derivatives −F indingM aximumP rofit
Problem Statement: A company produces and sells handmade candles. The
weekly profit function Pin dollars, as a function of the number of candles x
sold, is given by:
P(x) = −0.1x2+ 40x−300
To maximize their profit, the company wants to determine the number of
candles they should produce and sell each week.
Questions:
a) Find the derivative of the profit function, P(x).
b) Determine the critical point(s) for the profit function.
c) Use the second derivative test to confirm whether the critical point is a
maximum.
d) What is the maximum weekly profit, and how many candles must be sold
to achieve this profit?
Solution:
a) Finding the derivative of the profit function, P(x):
Step-by-Step:
1. Given Function:
P(x) = −0.1x2+ 40x−300
2. Differentiate P(x) with respect to x:
P′(x) = d
dx (−0.1x2+ 40x−300)
P′(x) = −0.2x+ 40
b) Determining the critical point(s):
Step-by-Step:
1. Set the derivative P′(x) to zero:
−0.2x+ 40 = 0
2. Solve for x:
−0.2x=−40
x= 200
c) Using the second derivative test:
Step-by-Step:
1. Find the second derivative P′′ (x):
P′′ (x) = d
dx (−0.2x+ 40)
13
P′′ (x) = −0.2
2. Evaluate P′′ (x) at the critical point x= 200:
P′′ (200) = −0.2
Since P′′ (200) <0, the function P(x) has a local maximum at x= 200.
d) Maximum weekly profit and number of candles:
Step-by-Step:
1. Substitute x= 200 into P(x) to find maximum profit:
P(200) = −0.1(200)2+ 40(200) −300
P(200) = −0.1(40000) + 8000 −300
P(200) = −4000 + 8000 −300
P(200) = 3700
Thus, the maximum weekly profit is 3700, andtoachievethis, thecompanymustproduceandsell200candleseachweek.
Question 7
Problem: A company manufactures a product that has a demand equation
given by p= 300−2q, where pis the price per unit in dollars and qis the quantity
sold per month. Find the rate at which revenue is changing with respect to price
when the price is 100.
Solution:
Step 1: Write the revenue function. The revenue Ris the product of the
price pand the quantity q. From the given demand equation p= 300 −2q, we
can solve for qin terms of p:
q=300 −p
2
The revenue function Rcan then be expressed as:
R=p·q=p·300 −p
2
R=300p−p2
2
Step 2: Differentiate the revenue function with respect to p. To find how
the revenue changes with respect to price, we differentiate Rwith respect to p:
dR
dp =d
dp 300p−p2
2
dR
dp =1
2(300 −2p)
14
dR
dp = 150 −p
Step 3: Evaluate dR
dp when p= 100. Substituting p= 100 into the derivative:
dR
dp = 150 −100
dR
dp = 50
Conclusion: The rate at which revenue is changing with respect to price
when the price is 100is50 per dollar increase in price. This means that if the
price per unit increases by 1, therevenuewillincreasebyapproximately50 when
the price is at 100.Question7 : Applicationsof Derivatives
Problem: A company manufactures a product that has a demand equation
given by p= 300−2q, where pis the price per unit in dollars and qis the quantity
sold per month. Find the rate at which revenue is changing with respect to price
when the price is 100.
Solution:
Step 1: Write the revenue function. The revenue Ris the product of the
price pand the quantity q. From the given demand equation p= 300 −2q, we
can solve for qin terms of p:
q=300 −p
2
The revenue function Rcan then be expressed as:
R=p·q=p·300 −p
2
R=300p−p2
2
Step 2: Differentiate the revenue function with respect to p. To find how
the revenue changes with respect to price, we differentiate Rwith respect to p:
dR
dp =d
dp 300p−p2
2
dR
dp =1
2(300 −2p)
dR
dp = 150 −p
Step 3: Evaluate dR
dp when p= 100. Substituting p= 100 into the derivative:
dR
dp = 150 −100
dR
dp = 50
15
Conclusion: The rate at which revenue is changing with respect to price
when the price is 100is50 per dollar increase in price. This means that if the
price per unit increases by 1, therevenuewillincreasebyapproximately50 when
the price is at 100.
Question 8
Problem Statement: A company produces and sells handmade candles. The
price per candle that consumers are willing to pay decreases as the quantity of
candles sold increases, according to the demand function:
p(x) = 20 −0.5x
where pis the price in dollars per candle and xis the number of candles sold
per day. The cost per day, C(x), of producing xcandles is given by:
C(x) = 100 + 3x
Find the number of candles the company should produce and sell each day
to maximize profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Determine the Revenue Function. Revenue, R, is the product of the
number of units sold, x, and the price per unit, p(x). So,
R(x) = x·p(x) = x(20 −0.5x)
R(x) = 20x−0.5x2
Step 2: Determine the Profit Function. Profit, Π, is the difference between
Revenue and Cost. So,
Π(x) = R(x)−C(x)
Π(x) = (20x−0.5x2)−(100 + 3x)
Π(x) = 20x−0.5x2−100 −3x
Π(x) = −0.5x2+ 17x−100
Step 3: Calculate the Derivative of the Profit Function. To find the profit
maximization point, take the derivative of Π(x) with respect to x, and set it to
zero.
Π′(x) = −x+ 17
Set Π′(x) = 0:
−x+ 17 = 0
x= 17
16
Step 4: Verify Maximum Profit Using Second Derivative Test. Compute the
second derivative of Π(x):
Π′′ (x) = −1
Since Π′′ (x) = −1 is negative, the function has a local maximum at x= 17.
Step 5: Calculate Maximum Profit. Plug x= 17 back into the profit function
to find the maximum profit.
Π(17) = −0.5(17)2+ 17(17) −100
Π(17) = −0.5×289 + 289 −100
Π(17) = −144.5 + 289 −100
Π(17) = 44.5
Conclusion: The company should produce and sell 17 candles per day to
maximize their profit, which will be 44.5perday.Question8 : Applicationsof Derivatives−
MaximumP rofitP roblem
Problem Statement: A company produces and sells handmade candles. The
price per candle that consumers are willing to pay decreases as the quantity of
candles sold increases, according to the demand function:
p(x) = 20 −0.5x
where pis the price in dollars per candle and xis the number of candles sold
per day. The cost per day, C(x), of producing xcandles is given by:
C(x) = 100 + 3x
Find the number of candles the company should produce and sell each day
to maximize profit. What is the maximum profit?
Step-by-Step Solution:
Step 1: Determine the Revenue Function. Revenue, R, is the product of the
number of units sold, x, and the price per unit, p(x). So,
R(x) = x·p(x) = x(20 −0.5x)
R(x) = 20x−0.5x2
Step 2: Determine the Profit Function. Profit, Π, is the difference between
Revenue and Cost. So,
Π(x) = R(x)−C(x)
Π(x) = (20x−0.5x2)−(100 + 3x)
Π(x) = 20x−0.5x2−100 −3x
Π(x) = −0.5x2+ 17x−100
17
Step 3: Calculate the Derivative of the Profit Function. To find the profit
maximization point, take the derivative of Π(x) with respect to x, and set it to
zero.
Π′(x) = −x+ 17
Set Π′(x) = 0:
−x+ 17 = 0
x= 17
Step 4: Verify Maximum Profit Using Second Derivative Test. Compute the
second derivative of Π(x):
Π′′ (x) = −1
Since Π′′ (x) = −1 is negative, the function has a local maximum at x= 17.
Step 5: Calculate Maximum Profit. Plug x= 17 back into the profit function
to find the maximum profit.
Π(17) = −0.5(17)2+ 17(17) −100
Π(17) = −0.5×289 + 289 −100
Π(17) = −144.5 + 289 −100
Π(17) = 44.5
Conclusion: The company should produce and sell 17 candles per day to
maximize their profit, which will be 44.5perday.
Question 9
A company produces and sells an item. The price per item, p, in dollars,
can be expressed as a function of the number of items, x, sold and is given by
the equation p(x) = 40 −0.1x. The revenue Ris the product of the number of
items sold and the price per item. The task is to find how many items should
be sold to maximize revenue.
Step-by-step Solution:
Step 1: Identify the revenue function Since revenue Ris the product of the
number of items sold, x, and the price per item, p(x), you first write down the
expression for Rusing the function for p(x):
R(x) = x×p(x) = x×(40 −0.1x)
R(x) = 40x−0.1x2
Step 2: Take the derivative of the revenue function Differentiate R(x) with
respect to xto find R′(x):
R′(x) = 40 −0.2x
18
Step 3: Set the first derivative equal to zero to find critical points Set R′(x) =
0 and solve for x:
40 −0.2x= 0
0.2x= 40
x= 200
Step 4: Evaluate the concavity using the second derivative or critical points
To ensure this value of xgives a maximum, check the sign of the second deriva-
tive, R′′ (x):
R′′ (x) = −0.2
Since R′′ (x)<0, the function is concave down at x= 200, indicating a local
maximum.
Step 5: Conclusion The company will maximize revenue when they sell 200
items. The maximum revenue can also be calculated:
R(200) = 40(200) −0.1(200)2= 8000 −4000 = 4000
Hence, the maximum revenue is 4000.
This provides a guided solution to maximizing the revenue for the company
based on the given price function. Question 9: Maximizing Revenue
A company produces and sells an item. The price per item, p, in
dollars, can be expressed as a function of the number of items, x, sold
and is given by the equation p(x) = 40 −0.1x. The revenue Ris the
product of the number of items sold and the price per item. The task
is to find how many items should be sold to maximize revenue.
Step-by-step Solution:
Step 1: Identify the revenue function Since revenue Ris the prod-
uct of the number of items sold, x, and the price per item, p(x), you
first write down the expression for Rusing the function for p(x):
R(x) = x×p(x) = x×(40 −0.1x)
R(x) = 40x−0.1x2
Step 2: Take the derivative of the revenue function Differentiate
R(x)with respect to xto find R′(x):
R′(x) = 40 −0.2x
Step 3: Set the first derivative equal to zero to find critical points
Set R′(x) = 0 and solve for x:
40 −0.2x= 0
0.2x= 40
x= 200
19
Step 4: Evaluate the concavity using the second derivative or crit-
ical points To ensure this value of xgives a maximum, check the sign
of the second derivative, R′′ (x):
R′′ (x) = −0.2
Since R′′ (x)<0, the function is concave down at x= 200, indicating a
local maximum.
Step 5: Conclusion The company will maximize revenue when they
sell 200 items. The maximum revenue can also be calculated:
R(200) = 40(200) −0.1(200)2= 8000 −4000 = 4000
Hence, the maximum revenue is 4000.
This provides a guided solution to maximizing the revenue for the
company based on the given price function.
Question 10
Problem Statement: A company produces and sells handmade bracelets.
The price per bracelet can be modeled by the demand function p(x) = 20−0.5x,
where pis the price in dollars and xis the number of bracelets sold per day.
The daily cost to produce xbracelets is given by C(x) = 40 + 2x. Determine
the number of bracelets the company should produce and sell to maximize its
daily revenue.
Solution:
Step 1: Define the Revenue Function The revenue function R(x) is defined as
the product of the number of units sold and the price per unit. Mathematically,
this is expressed as:
R(x) = x·p(x)
Substituting the demand function:
R(x) = x·(20 −0.5x) = 20x−0.5x2
Step 2: Compute the Derivative of the Revenue Function To find the maxi-
mum revenue, we need to find the critical points of R(x). This is done by taking
the derivative of R(x) and setting it to zero:
R′(x) = 20 −x
Set R′(x) = 0 to find the critical points:
20 −x= 0 =⇒x= 20
Step 3: Determine Whether This Critical Point Is a Maximum To confirm
whether this critical point x= 20 is a maximum, use the second derivative test.
Compute R′′ (x):
R′′ (x) = −1
20
Since R′′ (20) = −1 is negative, the function has a local maximum at x= 20.
Step 4: Evaluate the Maximum Revenue Plug x= 20 back into the revenue
function to find the maximum revenue:
R(20) = 20 ×20 −0.5×202= 400 −200 = 200
So, the maximum revenue is 200.
Step 5: Conclusion The company should produce and sell 20 bracelets daily
to maximize its revenue, which will be 200perdayatthislevelof production.
This answer outlines how the applications of derivatives, specifically using
the first and second derivative tests, are crucial in real-world business scenarios
like maximizing revenue. Question 10: Maximum Revenue Problem
Problem Statement: A company produces and sells handmade
bracelets. The price per bracelet can be modeled by the demand
function p(x) = 20 −0.5x, where pis the price in dollars and xis
the number of bracelets sold per day. The daily cost to produce x
bracelets is given by C(x) = 40+2x. Determine the number of bracelets
the company should produce and sell to maximize its daily revenue.
Solution:
Step 1: Define the Revenue Function The revenue function R(x)is
defined as the product of the number of units sold and the price per
unit. Mathematically, this is expressed as:
R(x) = x·p(x)
Substituting the demand function:
R(x) = x·(20 −0.5x) = 20x−0.5x2
Step 2: Compute the Derivative of the Revenue Function To find
the maximum revenue, we need to find the critical points of R(x).
This is done by taking the derivative of R(x)and setting it to zero:
R′(x) = 20 −x
Set R′(x) = 0 to find the critical points:
20 −x= 0 =⇒x= 20
Step 3: Determine Whether This Critical Point Is a Maximum
To confirm whether this critical point x= 20 is a maximum, use the
second derivative test. Compute R′′ (x):
R′′ (x) = −1
Since R′′ (20) = −1is negative, the function has a local maximum at
x= 20.
21
Step 4: Evaluate the Maximum Revenue Plug x= 20 back into the
revenue function to find the maximum revenue:
R(20) = 20 ×20 −0.5×202= 400 −200 = 200
So, the maximum revenue is 200.
Step 5: Conclusion The company should produce and sell 20 bracelets
daily to maximize its revenue, which will be 200perdayatthislevelof production.
This answer outlines how the applications of derivatives, specifi-
cally using the first and second derivative tests, are crucial in real-
world business scenarios like maximizing revenue.
22
