Revenue Functions
In general, a business is concerned not only with its costs, but also with its revenues. Recall that, if R(x) is the revenue received from the sale of x units of some commodity, then the derivative R(x) is called the marginal revenue. Economists use this to measure the rate of increase in revenue per unit increase in sales. If x units of a product are sold at a price p per unit, the total revenue R(x) is given by R(x) = x· p.
If a firm is small and is in competition with many other companies, its sales have little effect on the market price. Then, since the price is constant as far as the one firm is concerned, the marginal revenue R(x) equals the price p that is, R(x) is the amount that the firm receives from the sale of one additional unit. In this case, the revenue function will have a graph as in Fig. 4.
If a firm is small and is in competition with many other companies, its sales have little effect on the market price. Then, since the price is constant as far as the one firm is concerned, the marginal revenue R(x) equals the price p that is, R(x) is the amount that the firm receives from the sale of one additional unit. In this case, the revenue function will have a graph as in Fig. 4.
An interesting problem arises when a single firm is the only supplier of a certain product or service, that is, when the firm has a monopoly. Consumers will buy large amounts of the commodity if the price per unit is low and less if the price is raised.
For each quantity x, let f(x) be the highest price per unit that can be set to sell all x units to customers. Since selling greater quantities requires a lowering of the price, f(x) will be a decreasing function. Figure 5 shows a typical demand curve that relates the quantity demanded, x, to the price, p = f(x).
For each quantity x, let f(x) be the highest price per unit that can be set to sell all x units to customers. Since selling greater quantities requires a lowering of the price, f(x) will be a decreasing function. Figure 5 shows a typical demand curve that relates the quantity demanded, x, to the price, p = f(x).
![Picture](/uploads/5/5/0/3/55032007/717592.png?250)
The demand equation p = f(x) determines the total revenue function. If the firm wants to sell x units, the highest price it can set is f(x) dollars per unit, and so the total revenue from the sale of x units is
R(x) = x·p = x·f(x).
The concept of a demand curve applies to an entire industry (with many producers)as well as to a single monopolistic firm. In this case, many producers offer the same product for sale. If x denotes the total output of the industry, f(x) is the market price per unit of output and x · f(x) is the total revenue earned from the sale of the x units.
R(x) = x·p = x·f(x).
The concept of a demand curve applies to an entire industry (with many producers)as well as to a single monopolistic firm. In this case, many producers offer the same product for sale. If x denotes the total output of the industry, f(x) is the market price per unit of output and x · f(x) is the total revenue earned from the sale of the x units.
EXAMPLE 1
The WMA Bus Lines offers sightseeing tours of Washington, D.C. One tour, priced at $7 per person, had an average demand of about 1000 customers per week. When the price was lowered to $6, the weekly demand jumped to about 1200 customers. Assuming that the demand equation is linear, find the tour price that should be charged per person to maximize the total revenue each week.
Using R(x) and R’(x), we can sketch the graph of R(x). (See Fig. 8.) The maximum revenue occurs when the marginal revenue is zero, that is, when x = 1200. The price corresponding to this number of customers is found from demand equation:
THE COST FUNCTION
The cost function is the function that relates the total cost C(x) of producing an x number of units of a product to that number x
Formula:
C(x) = p(x) * x
Formula:
C(x) = p(x) * x
MARGINAL COST
The derivative of C is also called the marginal cost.
It is the approximate increase in cost of producing one more item.
FORMULA:
C’(x)
It is the approximate increase in cost of producing one more item.
FORMULA:
C’(x)
EXAMPLE 2
C(x) = 450 + 30x – 9 + 4
= 0 + 30 – 2(9x) + 3(4 )
= 30 – 18x + 12
=12x^2 – 18x + 30
= 0 + 30 – 2(9x) + 3(4 )
= 30 – 18x + 12
=12x^2 – 18x + 30
EXAMPLE 3
The demand equation for a certain product is
P= 12 – ½ x dollars
Find the level of production that maximizes revenue.
We want to begin by developing our revenue function R(x). R(x) = X so,
R(x) = x.p = x ( 12 – ½ x) =12x – ½ x²
The marginal revenue is given by R’(x)
R’ (x) = 12 – x
We use R’(x) = 0 to identify the relative maximum point.
12 –x = 0
x = 12
The corresponding value of y is
R(12)= 12.12- ½(12)²= 72 dollars.
P= 12 – ½ x dollars
Find the level of production that maximizes revenue.
We want to begin by developing our revenue function R(x). R(x) = X so,
R(x) = x.p = x ( 12 – ½ x) =12x – ½ x²
The marginal revenue is given by R’(x)
R’ (x) = 12 – x
We use R’(x) = 0 to identify the relative maximum point.
12 –x = 0
x = 12
The corresponding value of y is
R(12)= 12.12- ½(12)²= 72 dollars.
THE PROFIT FUNCTION
The profit function is nothing but the revenue function minus the cost function
So the formula is:
P(x) = R(x)-C(x)
→x.p(x)-C(x)
Notice that:
A maximum profit is reached when:
1. The first derivative of P(x) when P‘(x) is zero or doesn’t exist
And
2. The second derivative of P(x) is always negative P″(x) < 0
*when P‘(x) = 0 this is the critical point, and since P″(x) < 0 the parabola will be concave downward
Another important notice is:
Since P(x) = R(x) - C(x)
Then a maximum profit can be reached when:
R'(x)-C'(x) = 0 and R″ (x)-C″(x) < 0
So the formula is:
P(x) = R(x)-C(x)
→x.p(x)-C(x)
Notice that:
A maximum profit is reached when:
1. The first derivative of P(x) when P‘(x) is zero or doesn’t exist
And
2. The second derivative of P(x) is always negative P″(x) < 0
*when P‘(x) = 0 this is the critical point, and since P″(x) < 0 the parabola will be concave downward
Another important notice is:
Since P(x) = R(x) - C(x)
Then a maximum profit can be reached when:
R'(x)-C'(x) = 0 and R″ (x)-C″(x) < 0
EXAMPLE 4
A factory that produces i-pods computed its demand, and costs and came to realize that the formulas are as follows:
p=100-0.01x
C(x) = 50x+10000
Find:
The number of units that should be produced for the factory to obtain maximum profit.
The price of the unit.
R(x) = x.p
=x.(100-0.01)
=100x -0.01x2
Now:
P(x) = R(x)-C(x)
=100x-0.01x2 – (50x+10000)
=-0.01x2 + 50x-10000
Now the maximum point in the parabola will occur When P'(x) = 0 ; P"(x)<0
P(x) = -0.01x2 + 50x-10000
P'(x) = -0.02x+50
→ -0.02x+50=0
→ -0.02x=-50
→ x =2500
P'(2500) = -50+50=0
So the P'(X) =0 When x = 2500
- Now we will find the second derivative:
P″(x) = -0.02
P″(2500)= -0.02<0
So x=2500 is at a local maximum
p=100-0.01x
C(x) = 50x+10000
Find:
The number of units that should be produced for the factory to obtain maximum profit.
The price of the unit.
R(x) = x.p
=x.(100-0.01)
=100x -0.01x2
Now:
P(x) = R(x)-C(x)
=100x-0.01x2 – (50x+10000)
=-0.01x2 + 50x-10000
Now the maximum point in the parabola will occur When P'(x) = 0 ; P"(x)<0
P(x) = -0.01x2 + 50x-10000
P'(x) = -0.02x+50
→ -0.02x+50=0
→ -0.02x=-50
→ x =2500
P'(2500) = -50+50=0
So the P'(X) =0 When x = 2500
- Now we will find the second derivative:
P″(x) = -0.02
P″(2500)= -0.02<0
So x=2500 is at a local maximum
GRAPH:
![Picture](/uploads/5/5/0/3/55032007/5720186.png?335)
Finally to find the price that is needed to be charged per unit we return to the demand function:
P=100-0.01(2500)
= 100-25= $75
P=100-0.01(2500)
= 100-25= $75
EXAMPLE 5
Ex:The demand function for a product is modeled by
p= 400 – x, 0 ≤ x ≤ 400
Where p is the price per unit and x is the number of units. Use differentials to approximate the change in revenue as sales increase from 149 units to 150 units. Compare this with actual change in revenue.
Solution: Begin by finding the revenue function. Because the demand is given by p = 400 – x, the revenue is
R = xp
= x(400 – x)
= 400x – x^2
Next find the marginal revenue, dR/dx which is 400 – 2x
When x = 149 and dx= Δx = 1, the approx. change in the revenue is
ΔR = dR
= (400 – 2x) dx
= [400 – 2(149)] (1) = $102.
When x increases from 149 to 150 and R = f(x) = 400x – x^2 , the actual change in revenue is
ΔR = f(x + Δx) – f(x)
= [400(150) – 1502] – [400(149) – 1492 = $101.
p= 400 – x, 0 ≤ x ≤ 400
Where p is the price per unit and x is the number of units. Use differentials to approximate the change in revenue as sales increase from 149 units to 150 units. Compare this with actual change in revenue.
Solution: Begin by finding the revenue function. Because the demand is given by p = 400 – x, the revenue is
R = xp
= x(400 – x)
= 400x – x^2
Next find the marginal revenue, dR/dx which is 400 – 2x
When x = 149 and dx= Δx = 1, the approx. change in the revenue is
ΔR = dR
= (400 – 2x) dx
= [400 – 2(149)] (1) = $102.
When x increases from 149 to 150 and R = f(x) = 400x – x^2 , the actual change in revenue is
ΔR = f(x + Δx) – f(x)
= [400(150) – 1502] – [400(149) – 1492 = $101.
MARGINAL COST
Marginal cost (MC) is the extra, or additional, cost of producing one more unit of output. MC can be determined for each added unit of output by noting the change in total cost that unit’s production entails:
MC= (Change in Total Cost)/ (Change in quantity)
MC= (Change in Total Cost)/ (Change in quantity)
EXAMPLE 6
Suppose the cost in dollars to manufacture portable music players is given by
C (x) = 150,000 + 20x – 0.0001x^2
Where x is the number of music players manufactured.
a. Find the average cost per music player if 50,000 music players are manufactured.
b. Find a formula for the average cost per music player if x music players is manufactured. This function of x is called the average cost function.
Solution:
a. The total cost of manufacturing 50,000 music players is given by
C(50,000)=150,000+ 20(50,000) –
0.0001(50,000)2 = $900,000
C(50,000)= 900,000/50,000 = $18.00
b. Ċ (x)= C (x)/x
= (1/x)(150,000+ 20x- 0.0001x2)
= (150,000/x) + 20 – 0.0001x
C (x) = 150,000 + 20x – 0.0001x^2
Where x is the number of music players manufactured.
a. Find the average cost per music player if 50,000 music players are manufactured.
b. Find a formula for the average cost per music player if x music players is manufactured. This function of x is called the average cost function.
Solution:
a. The total cost of manufacturing 50,000 music players is given by
C(50,000)=150,000+ 20(50,000) –
0.0001(50,000)2 = $900,000
C(50,000)= 900,000/50,000 = $18.00
b. Ċ (x)= C (x)/x
= (1/x)(150,000+ 20x- 0.0001x2)
= (150,000/x) + 20 – 0.0001x
EXAMPLE 7
The Pocket EZCie is a miniature key chain flashlight based on led technology. The cost function is
8^(3/4) + 300
Dollars, where x is the number of pocket EZCies produced.
a. Find the marginal cost function MC (x)
b. Find the marginal cost when 81 Pocket EZCIes have been produced and interpret your answer.
Solution:
a. MC (x) = derivative of the cost function.
MC (x) = 6x^(-1/4)
b. To find the marginal cost when 81 Pocket EZCies have been produced, we evaluate the marginal cost function MC (x) at x= 81:
MC (81)= 6/81^(-1/4) = 6/3 = 2
Interpretation: When 81 Pocket EZCies have been produced, the marginal cost is $2, meaning that to produce one more Pocket EZCie costs about $2.
8^(3/4) + 300
Dollars, where x is the number of pocket EZCies produced.
a. Find the marginal cost function MC (x)
b. Find the marginal cost when 81 Pocket EZCIes have been produced and interpret your answer.
Solution:
a. MC (x) = derivative of the cost function.
MC (x) = 6x^(-1/4)
b. To find the marginal cost when 81 Pocket EZCies have been produced, we evaluate the marginal cost function MC (x) at x= 81:
MC (81)= 6/81^(-1/4) = 6/3 = 2
Interpretation: When 81 Pocket EZCies have been produced, the marginal cost is $2, meaning that to produce one more Pocket EZCie costs about $2.
EXAMPLE 8
Each council of girl scouts has costs associated with acquiring and selling cookies during their annual cookie campaign. The council sets the price per box sold in their region. During a recent campaign, one council set the sales price for their region at $4.00 per box. The total cost to the council associated with the campaign can be modeled as
C(q) = 0.23q3 – 0.98q2 + 2.7q + 0.2 million dollars
When q million boxes were sold, 0 < q < 4.
a. Find the models for marginal cost, revenue, and marginal revenue with respect to q number of boxes (in millions)
b. What is the cost for the 2,000,001st box? What is the revenue for that box?
c. How many boxes should the council sell to maximize profit? What is the profit for this level of sales?
Solutions:
a. MC is the derivative of cost
C(q) = 0.69q2 – 1.96q + 2.7 dollars per box
Revenue is the selling price times the number sold:
R(q)= 4 dollars per box
Marginal Revenue is the derivative of revenue:
R’(q) = 4 dollars per box
b. The 2,000,001st box of cookies costs the council C’ (2) = 1.54 dollars and will bring in revenue of R’ (2) = 4 dollars
c. According to the profit maximization rule, profit is maximized at an input level for which marginal cost equals marginal revenue. Solving
0.69q2 – 1.96q + 2.7 = 4
Yields q= 3.395 million boxes. Profit from this level of sales is
P(3.395)= R(3.395) – C(3.395) = 6.509 million dollars
C(q) = 0.23q3 – 0.98q2 + 2.7q + 0.2 million dollars
When q million boxes were sold, 0 < q < 4.
a. Find the models for marginal cost, revenue, and marginal revenue with respect to q number of boxes (in millions)
b. What is the cost for the 2,000,001st box? What is the revenue for that box?
c. How many boxes should the council sell to maximize profit? What is the profit for this level of sales?
Solutions:
a. MC is the derivative of cost
C(q) = 0.69q2 – 1.96q + 2.7 dollars per box
Revenue is the selling price times the number sold:
R(q)= 4 dollars per box
Marginal Revenue is the derivative of revenue:
R’(q) = 4 dollars per box
b. The 2,000,001st box of cookies costs the council C’ (2) = 1.54 dollars and will bring in revenue of R’ (2) = 4 dollars
c. According to the profit maximization rule, profit is maximized at an input level for which marginal cost equals marginal revenue. Solving
0.69q2 – 1.96q + 2.7 = 4
Yields q= 3.395 million boxes. Profit from this level of sales is
P(3.395)= R(3.395) – C(3.395) = 6.509 million dollars
EXAMPLE 9
The relative extreme point of P(x) will be the point at which profit is the highest. We need to compute the marginal profit function P’(x), and solve P’(x) = 0 for x.
P’ = - .01
P’(x) = -.02x + 50
Solving P’(x) = 0, we have
-.02(x-2500) = 0
Which yields x = 2,500
If 2,500 units are produced, the profit would be
P (2,500) = -.01(2,500^2) + 50 (2,500) – 10,000
=52,500 dollars
The highest price at which the 2,500 units can be sold is
P = 100 - .01(2,500)
= 100 - 25 = 75 dollars
Answer:
Produce 2,500 units and sell them at $75 per unit.
The profit will be $52,500.
P’ = - .01
P’(x) = -.02x + 50
Solving P’(x) = 0, we have
-.02(x-2500) = 0
Which yields x = 2,500
If 2,500 units are produced, the profit would be
P (2,500) = -.01(2,500^2) + 50 (2,500) – 10,000
=52,500 dollars
The highest price at which the 2,500 units can be sold is
P = 100 - .01(2,500)
= 100 - 25 = 75 dollars
Answer:
Produce 2,500 units and sell them at $75 per unit.
The profit will be $52,500.