1. ECONOMIC APPLICATIONS

2. EXAMPLE 1
Ardine makes and sells mukluks at a craft fair. Last year, she sold mukluks for $400 per pair, and she sold 14 pairs. She predicts that for every $40 increase in price, she will sell one fewer pair.
The revenue from the mukluk sales,
R(x) = (Number of Mukluks Sold)(Cost Per Mukluk)
Number of changes x
Number of Mukluks Sold
Cost per Mukluk
($)
Revenue, R(x) 14
400 1 2 3 4 5
5600
5600 13
440
5720
Maximum
Revenue 12
480
5760
5720 11
520
5600 10
560 9
600
5400

3. Let x represent the number of increases to her price
EXAMPLE 1 continued
Ardine makes and sells mukluks at a craft fair. Last year, she sold mukluks for $400 per pair, and she sold 14 pairs. She predicts that for every $40 increase in price, she will sell one fewer pair.
 The revenue from the mukluk sales, R(x) = (Number of Mukluks Sold)(Cost Per Mukluk)
Let x represent the number of increases to her price
R(x) =(14 – 1x)( x)
R'(x) = ( x)(–1) + (14 –1x)(40)
R'(x) = – 400 – 40x – 40x
R'(x) = 160 – 80x
0 = 160 – 80x
x = 2
Ardine should increase her price 2 times to (2) = $480.
She will then expect to sell 14 – 1(2) = 12 pair for
revenue of 𝟒𝟖𝟎×𝟏𝟐=$𝟓𝟕𝟔𝟎

4. R(x) = (100 – 5x)(8 + 2x) R'(x) = (8 + 2x)(–5) + (100 – 5x)(2)
Practice Question 1
A sporting goods store sells reusable sports water bottles for $8. At this price their weekly sales are approximately 100 items. Research says that for every $2 increase in price, the manager can expect the store to sell five fewer water bottles. What selling price will give the maximum revenue? What is the maximum revenue he can expect?
R(x) = (100 – 5x)(8 + 2x)
R'(x) = (8 + 2x)(–5) + (100 – 5x)(2)
R'(x) = 200 – 10x – 40 – 10x
R'(x) = 160 – 20x
0 = 160 – 20x
x = 8
The store should increase their price 8 times to 8 + 2(8) = $24.
They will then expect to sell 100 – 5(8) = 60 bottles for
revenue of $𝟐𝟒×𝟔𝟎=$𝟏𝟒𝟒𝟎

5. R(x) = (200 + 50x)(70 – 1x) R'(x) = (70 – 1x)(50) + (200 + 50x)(–1)
Practice Question 2
A concert promoter is planning the ticket price for an upcoming concert for a certain band. At the last concert, she charged $70 per ticket and sold 200 tickets After conducting a survey, the promoter has determined that or every $1 decrease in ticket price she might expect to sell 50 more tickets.
R(x) = ( x)(70 – 1x)
R'(x) = (70 – 1x)(50) + ( x)(–1)
R'(x) = 3500 – 50x – 200 – 50x
R'(x) = 3300 – 100x
0 = 3300 – 100x
x = 33
The promoter should decrease her price 33 times to 70 – 1(33) = $37.
She will then expect to sell (33) = 1850 tickets for
revenue of $𝟑𝟕×𝟏𝟖𝟓𝟎=$𝟔𝟖 𝟒𝟓𝟎

6. Profit = # of bears x profit per bear – fixed cost
EXAMPLE 2
A toy retailer finds that 100 teddy bears can be sold if each is priced at $12. However, 20 more can be sold for every $1 decrease in price. The cost of each bear is $6. In addition, the retailer must pay a fixed cost of $150. How should the bears be priced in order to realize the most profit?
Profit = # of bears x profit per bear – fixed cost
Let x be the number of decreases to the price
Original Profit PER bear is $12 – $6 = $6
Profit = ( x)(6 – 1x) –150
P(x) = 600 – 100x + 120x – 20x2 – 150
P(x) = x – 20x2
P'(x) = – 40x + 20

7. New Selling Price will be $6 + $5.50 $11.50
EXAMPLE 2: Continued
P'(x) = – 40x + 20
0 = – 40x + 20
x = 0.5
Profit = 6 – 1(0.5) = $5.50
New Selling Price will be $6 + $5.50 $11.50

8. PRACTICE QUESTION 3 PAGE 4
Many years ago an apartment management firm tended an apartment block containing 150 units. All 150 units are rented at $460 per unit with each unit costing $72.50/month for utility and repairs. For every $25 rent increase, 4 fewer apartments are occupied. What rent should be charged in order to realize the most profit? (Assume that rent would only be increased by a multiple of $25.)
Profit PER unit = $460 – $72.50 = $387.50
Let x be the number of increases to the price
R(x) = (150 – 4 x)( x)
R(x) = x – 1550x – 100x2
R(x) = x – 100x2

9. PRACTICE QUESTION 3 Continued
R(x) = x – 100x2
R'(x) = 2200 – 200x
0 = 2200 – 200x
2200 = 200x
x = 11
150 – 4(11) = (11) = $662.50
New Rental Cost = $ $72.50 = $735.00

10. Let x represent the number of weeks to wait
EXAMPLE 3
If a farmer harvests his crop today he will have 1200 bushels worth $2.00 per bushel. Every week he waits, the crop increases by 100 bushels, but the price drops 10 cents per bushel. When should he harvest the crop?
Let x represent the number of weeks to wait
P(x) = ( x)(2 – 0.1x)
P'(x) = (2 – 0.1x)(100) + ( x)(0.1)
P'(x) = 200 – 10x – 120 – 10x
P'(x) = 80 – 20x
0 = 80 – 20x
x = 4
The farmer should wait 4 weeks when he has (4) = 1600 bushels and sells at 2 – 0.1(4) = $ This will give him a maximum revenue of $2560.

11. Practice Question 3
Luigi owns a potato farm in southern Alberta. Each year he faces a dilemma as to when to harvest his crop of potatoes. From past experience, he knows that if he harvests on July 15, he can expect approximately 2000 kg of potatoes which he could sell at $0.60 per kg. For each week he waits after July 15, he can expect an extra 400 kg of potatoes, but the price will reduce by $0.50 per kg. When should he harvest his crop for maximum revenue?
R(x) = ( x)(0.60 – 0.05x)
R'(x) = (0.60 – 0.05x)(400) + ( x)(–0.05)
R'(x) = 240 – 20x – 100 – 20x
R'(x) = 140 – 40x
0 = 140 – 40x
x = 3.5
Luigi should wait 3.5 weeks after July 15 to harvest his crop.

12. For those interested in an optional more in-depth look at some economic problems using Calculus should study the following slides. Not all economic problems can be represented by simple quadratic functions as we have previously looked at.
I will not be testing you on the definitions and approaches covered in the following slides.

13. Economic Function Definitions
Cost Function 𝑪(𝒙): what it costs a company to produce 𝑥 units of a certain
commodity.
Marginal Cost : 𝐶 ′ 𝑥 𝑜𝑟 𝑑𝐶 𝑑𝑥
Demand (price) Function 𝒑(𝒙): price per unit sold
Revenue Function: If x units are sold and the price per unit is 𝒑(𝒙), then the
total revenue is 𝑹 𝒙 =𝒙𝒑(𝒙)
Marginal Revenue Function 𝑹′(𝒙): the rate of change of revenue with
respect to the number of units sold.
Profit Function 𝑷 𝒙 =𝑹 𝒙 −𝑪(𝒙)
Marginal Profit Function 𝑷′(𝒙) the derivative of the profit function.

14. When selling a product, total revenue or income depends on the selling price of the product and the number of units sold. The price of a product may be determined by the number of units sold. If a product is selling well, the price per unit may decrease.
If it costs a company C(x) dollars to produce x units of a product, then the function defined by C is called the
cost function.
The total cost of producing x units of a product includes fixed costs (rent, utilities, depreciation on machinery etc.), which continue even if nothing is produced, and variable costs (labour, materials, etc.), which depend on the number of items produced.

15. Cost function = fixed costs + variable costs C(x) = F + p(x)
In the cost function, x is the number of units produced and p(x) is the price per unit.
Economists call the derivative of the cost function the marginal cost. This derivative is used to calculate minimum costs.
The average cost function
𝑐 𝑥 = 𝐶(𝑥) 𝑥 ,
calculates the cost per unit when x units are produced.
When the average cost is at a minimum, it equals marginal cost

16. Total cost = fixed costs + variable costs 𝑪 𝒙 =𝟐𝟎 𝟎𝟎𝟎+𝒙 𝟎.𝟓𝒙+𝟏𝟐𝟎
Example 1
A manufacturing company determines that the variable cost of producing a certain number of microwave ovens is described by the function 𝒙 𝟎.𝟓𝒙+𝟏𝟐𝟎 . If the fixed costs are calculated to be $20 000, how many microwave ovens should be produced in order for the average costs to be minimized?
Let x be the number of microwave ovens to be manufactured.
Let c be the average cost to be minimized.
Total cost = fixed costs + variable costs
𝑪 𝒙 =𝟐𝟎 𝟎𝟎𝟎+𝒙 𝟎.𝟓𝒙+𝟏𝟐𝟎
𝑪 𝒙 =𝟐𝟎 𝟎𝟎𝟎+𝟎.𝟓 𝒙 𝟐 +𝟏𝟐𝟎𝒙

17. The average cost function
𝑐 𝑥 = 𝐶(𝑥) 𝑥 = 𝑥 𝑥 𝑥
𝑐(𝑥)= 𝑥 +0.5𝑥+120
𝑐 𝑥 = 𝑥 −1 +0.5𝑥+120
The minimum average cost can be found in two ways:
Method 1
Find x when the derivative of the average cost is 0.
𝒅𝒄 𝒅𝒙 =−𝟐𝟎 𝟎𝟎𝟎 𝒙 −𝟐 +𝟎.𝟓
𝟎=− 𝟐𝟎 𝟎𝟎𝟎 𝒙 𝟐 +𝟎.𝟓
𝟐𝟎 𝟎𝟎𝟎 𝒙 𝟐 =𝟎.𝟓
𝟐𝟎 𝟎𝟎𝟎=𝟎.𝟓 𝒙 𝟐
𝒙 𝟐 =𝟒𝟎 𝟎𝟎𝟎
𝒙=𝟐𝟎𝟎

18. Method 2
When marginal cost = average cost, the average cost will be a minimum.
The derivative of the cost function is marginal cost.
𝑪 𝒙 =𝟐𝟎 𝟎𝟎𝟎+𝟎.𝟓 𝒙 𝟐 +𝟏𝟐𝟎𝒙
𝒅𝑪 𝒅𝒙 =𝒙+𝟏𝟐𝟎
𝒙+𝟏𝟐𝟎= 𝟐𝟎 𝟎𝟎𝟎 𝒙 +𝟎.𝟓𝒙+𝟏𝟐𝟎
𝟎.𝟓𝒙= 𝟐𝟎 𝟎𝟎𝟎 𝒙
𝒙 𝟐 =𝟒𝟎 𝟎𝟎𝟎
𝒙=𝟐𝟎𝟎
The average cost is a minimum if 200 microwaves are manufactured

19. Practice Question
Quinton Mills is a large producer of flour. Management estimates that the cost (in dollars) of producing 𝑥 5-kg bags of flour is
𝐶 𝑥 = 𝑥 𝑥 2
a) Find the average cost of producing bags.
𝒄 𝒙 = 𝑪(𝒙) 𝒙 = 𝟏𝟒𝟎 𝟎𝟎𝟎+𝟎.𝟒𝟑𝒙+ 𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏𝒙 𝟐 𝒙
𝒄 𝒙 = 𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 +𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏𝒙
𝒄 𝒙 =𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 −𝟏 +𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏𝒙

20. b) Find the marginal cost of producing 100 000 bags.
𝑪 ′ 𝒙 =𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟐𝒙
𝑪 ′ 𝟏𝟎𝟎 𝟎𝟎𝟎 =𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟐 𝟏𝟎𝟎 𝟎𝟎𝟎 =$𝟎.𝟔𝟑
c) At what production level will the average cost be smallest,
and what is this average cost?
Method 1
Find x when the derivative of the average cost is 0.
𝒄 𝒙 =𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 −𝟏 +𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏𝒙
𝒄 ′ 𝒙 =−𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 −𝟐 +𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏
𝟎=− 𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 𝟐 +𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏
𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 𝟐 =𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏
𝒙 𝟐 =𝟏𝟒𝟎 𝟎𝟎𝟎 𝟎𝟎𝟎 𝟎𝟎𝟎
𝒙≈𝟑𝟕𝟒 𝟎𝟎𝟎

21. When marginal cost = average cost, the average cost will be a minimum.
Method 2
When marginal cost = average cost, the average cost will be a minimum.
The derivative of the cost function is marginal cost.
𝑪 𝒙 =𝟏𝟒𝟎 𝟎𝟎𝟎+𝟎.𝟒𝟑𝒙+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏 𝒙 𝟐
𝒎𝒂𝒓𝒈𝒊𝒏𝒂𝒍 𝒄𝒐𝒔𝒕=𝑪 ′ 𝒙 =𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟐𝒙
𝒂𝒗𝒆𝒓𝒂𝒈𝒆 𝒄𝒐𝒔𝒕= 𝒄 𝒙 = 𝑪(𝒙) 𝒙 = 𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 +𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏𝒙
𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟐𝒙= 𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙 +𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏𝒙
𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏𝒙= 𝟏𝟒𝟎 𝟎𝟎𝟎 𝒙
𝒙 𝟐 =𝟏𝟒𝟎 𝟎𝟎𝟎 𝟎𝟎𝟎 𝟎𝟎𝟎=𝟏𝟒× 𝟏𝟎 𝟏𝟎
𝒙≈𝟑𝟕𝟒 𝟎𝟎𝟎
The average cost will be a minimum when the production level is about bags and this minimum average cost is
𝟏𝟒𝟎 𝟎𝟎𝟎 𝟏𝟒× 𝟏𝟎 𝟏𝟎 +𝟎.𝟒𝟑+𝟎.𝟎𝟎𝟎 𝟎𝟎𝟏 𝟏𝟒× 𝟏𝟎 𝟏𝟎 =$𝟏.𝟏𝟖

22. Example
A store has been selling 200 compact disc players a week at $350 each. A market survey indicates that for each $10 rebate offered to the buyers the number of set sold will increase by 20 a week.
a) Find the demand function and the revenue function
If 𝑥 is the number of CD players sold per week, then the weekly increase in sales is 𝒙−𝟐𝟎𝟎.
For each increase of 20 players sold the price is decreased by $10. So for each additional player sold the decrease in price will be 𝟏 𝟐𝟎 $𝟏𝟎 =$ 𝟏 𝟐
The demand (price) function is 𝒑 𝒙 =𝟑𝟓𝟎− 𝟏 𝟐 𝒙−𝟐𝟎𝟎 =𝟒𝟓𝟎− 𝟏 𝟐 𝒙
The revenue function is 𝑹 𝒙 =𝒙𝒑 𝒙 =𝒙 𝟒𝟓𝟎− 𝟏 𝟐 𝒙 =𝟒𝟓𝟎𝒙− 𝟏 𝟐 𝒙 𝟐

23. b) How large a rebate should the store offer to maximize its revenue?
𝑹 𝒙 =𝟒𝟓𝟎𝒙− 𝟏 𝟐 𝒙 𝟐
𝑹 ′ 𝒙 =𝟒𝟓𝟎−𝒙
𝟎=𝟒𝟓𝟎−𝒙
𝒙=𝟒𝟓𝟎
The price that will maximize revenue
𝒑 𝟒𝟓𝟎 =𝟒𝟓𝟎− 𝟏 𝟐 𝟒𝟓𝟎 =$𝟐𝟐𝟓
To maximize revenue the store should offer a rebate of
$𝟑𝟓𝟎 −$𝟐𝟐𝟓=$𝟏𝟐𝟓

24. Practice Question
A chain of stores has been selling a line of cameras for $50 each and has been averaging sales of 8000 cameras a month. They decide to increase the price, but their market research indicates that for each $1 increase in price, sales will fall by 100.
a) Find the demand function and the revenue function
If 𝑥 is the number of cameras sold per week, then the weekly decrease in sales is 𝟖𝟎𝟎𝟎−𝒙.
For each decrease of 100 players sold the price is increased by $1. So for each additional player sold the decrease in price will be 𝟏 𝟏𝟎𝟎 $𝟏 =$𝟎.𝟎𝟏
The demand (price) function is 𝒑 𝒙 =𝟓𝟎+𝟎.𝟎𝟏 𝟖𝟎𝟎𝟎−𝒙 =𝟏𝟑𝟎−𝟎.𝟎𝟏𝒙
The revenue function is 𝑹 𝒙 =𝒙𝒑 𝒙 =𝒙 𝟏𝟑𝟎−𝟎.𝟎𝟏𝒙 =𝟏𝟑𝟎𝒙−𝟎.𝟎𝟏 𝒙 𝟐

25. The price that will maximize revenue
b) Find the price that will maximize their revenue?
𝑹 𝒙 =𝟏𝟑𝟎𝒙−𝟎.𝟎𝟏 𝒙 𝟐
𝑹 ′ 𝒙 =𝟏𝟑𝟎−𝟎.𝟎𝟐𝒙
𝟎=𝟏𝟑𝟎−𝟎.𝟎𝟐𝒙
𝒙=𝟔𝟓𝟎𝟎
The price that will maximize revenue
𝒑 𝟔𝟓𝟎𝟎 =𝟏𝟑𝟎−𝟎.𝟎𝟏𝒙 𝟔𝟓𝟎𝟎 =$𝟔𝟓
To maximize revenue the store should charge $65.