Chapter 10 – 7b Marginal Analysis
Example 6. The price-demand equation and the cost function for the production of table saws are given,
respectively, by
𝑥 = 6,000 − 30𝑝 and 𝐶(𝑥) = 72,000 + 60𝑥
where x is the number of saws that can be sold at a price of p dollars per saw and C(x) is the total cost of producing
x saws.
A) Find the marginal cost.
𝐶 ′ (𝑥) = 60
B) Find the marginal average cost.
𝐶̅ (𝑥) =
𝐶(𝑥) 72,000 + 60𝑥 72,000
=
=
+ 60
𝑥
𝑥
𝑥
𝐶̅ ′(𝑥) = −
72,000
𝑥2
C) Find the marginal cost and the marginal average cost at a production
level of 2,000 saws and interpret the results.
Marginal Cost and
Marginal Average Cost
𝐶 ′ (2,000) = 60
𝐶
̅ ′ (2,000)
Marginal Cost
72,000
=−
= 0.018
2,0002
At a production level of 2,000 saws, the total cost is increasing at the
rate of $60 per saw while the average cost per saw is decreasing at the
rate of 1.8 cents per saw.
D) Find the revenue function and determine its domain.
From the price-demand equation, 𝑝 = 200 −
𝑅(𝑥) = 𝑥𝑝 = 200𝑥 −
𝑥2
30
0 ≤ 𝑥 ≤ 6,000
E) Find the marginal revenue.
𝑅′ (𝑥) = 200 −
𝑥
15
𝑥
30
0 ≤ 𝑥 ≤ 6,000
Marginal Average Cost
$80
$0.00
$60
-$0.40
$40
-$0.80
$20
-$1.20
$0
-$1.60
0
2000
4000
Number of Table Saws
6000
F) Find the marginal revenue for x = 1,500 and x = 4,500 and interpret the results
𝑅′ (1500) = 200 −
1500
= 100 The total revenue is increasing at the rate of $100 per table saw sold.
15
𝑅′ (4500) = 200 −
4500
= −100 The total revenue is decreasing at the rate of $100 per table saw sold.
15
G) Graph the cost function and the revenue function on the
same coordinate system. Find the break-even points and
indicate regions of profit and loss.
The break-even points are at x=600 and x=3600. Sales
between 600 and 3600 (exclusive) will generate a profit. Sales
less than 600 or greater than 3600 will generate a loss.
H) Find the profit function.
𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 200𝑥 −
𝑥2
𝑥2
− (72,000 + 60𝑥) = 140𝑥 −
− 72000
30
30
I) Find the marginal profit.
𝑃
′ (𝑥)
𝑥
= 140 −
15
Profit
$100,000
$0
J) Find the marginal profit for x = 1,500 and x = 3,000
and interpret the results.
0
1000
2000
3000
4000
5000
6000
-$100,000
-$200,000
𝑃
′ (1500)
= 140 −
1500
15
= 40, so the total profit is
increasing at the rate of $40 per saw.
𝑃′ (3000) = 140 −
3000
15
= −60, so the total profit is
-$300,000
-$400,000
-$500,000
Number of Table Saws Sold
decreasing at the rate of $60 per saw.
K) How many units should be sold to maximize the total profit?
The maximum profit is realized at the point where the marginal profit is zero (i.e., where the profit stops increasing
and starts decreasing):
𝑃′ (𝑥) = 0 ⟹ 140 −
𝑥
15
= 0 ⟹ 𝑥 = 2,100, so the maximum profit is P(2100) = $75,000.
Example 7. A company is planning to manufacture and market a new two-slice electric toaster. After conducting
extensive market surveys, the research department provides the following estimates: a weekly demand of 200
toasters at a price of $16 per toaster and a weekly demand of 300 toasters at a price of $14 per toaster. The
financial department estimates that weekly fixed costs will be $1,400 and variable costs (cost per toaster) will be
$4.
A) Assuming the relationship between price and demand is linear, find both the demand as a function of price and
the price as a function of demand.
𝑥 = 1000 − 50𝑝
𝑝 = 20 − 0.02𝑥
Break-Even Analysis
B) Plot the revenue and cost functions on the same
coordinate system. Identify break-even points and regions
of loss and profit.
$6,000
𝐶(𝑥) = 1400 + 4𝑥
$4,000
𝑅(𝑥) = 𝑥𝑝 = 20𝑥 − 0.02𝑥 2
The break-even points are at x = 100 and x = 700. Sales
between 100 and 700 units (exclusive) will result in a profit.
Sales below 100 or above 700 will result in a loss.
$5,000
$3,000
Cost
$2,000
Revenue
$1,000
$0
0
200
400
600
800
1000
Number of Toasters Sold
C) Plot the graph of the profit function. Find the marginal
profit at x = 250 and x = 475 and interpret the results.
𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = −1400 + 16𝑥 − 0.02𝑥 2
Total Profit
𝑃′ (𝑥) = 16 − 0.04𝑥
𝑃′ (250) = 16 − 0.04(250) = 6. At a sales level of 250 toasters,
the profit is increasing at the rate of $6.00 per toaster.
𝑃′ (475) = 16 − 0.04(475) = −3. At a sales level of 475 toasters,
the profit is decreasing at the rate of $3.00 per toaster.
D) How many toasters should be sold to maximize the total profit?
$3,000
$2,000
$1,000
$0
-$1,000 0
-$2,000
-$3,000
-$4,000
-$5,000
-$6,000
200
400
600
800
1000
Number of Toasters Sold
The maximum profit will be realized when the marginal profit is zero (when the profit stops increasing and starts
decreasing).
16 − 0.04𝑥 = 0 ⟹ 𝑥 = 400 and 𝑃(400) = $1,800/week
Example 8. The price-demand equation and the cost function for the production of garden hoses are given,
respectively, by
𝑝 = 20 − √𝑥 and 𝐶(𝑥) = 500 + 2𝑥
where x is the number of garden hoses sold per week, p is the price in dollars, and C(x) is the total cost to product x
garden hoses.
A) Graph the revenue and cost functions on the same
coordinate system, find the break-even points, and
indicate the regions of profit and loss.
𝑅(𝑥) = 𝑥𝑝 = 20𝑥 − 𝑥 √𝑥
The break-even points are at x = 44 and x = 258 (rounded
to the nearest integer). Sales of 44 to 257 (inclusive) units
will result in a profit. Sales of 43 or fewer units or 258 or
more units will result in a loss.
Break-Even Analysis
$1,400
$1,200
$1,000
$800
$600
Cost
$400
Revenue
$200
$0
0
100
200
300
400
Number of Garden Hoses Sold
B) How many garden hoses need to be made to maximize the
profit and what is the maximum profit?
Profit
$500
𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 18𝑥 − 𝑥 √𝑥 − 500
$0
3
𝑃′ (𝑥) = 18 − √𝑥 = 0 ⟹ 𝑥 = 144
2
⟹ Maximum profit is 𝑃(144) = $364/week.
0
50
100
150
200
250
300
-$500
-$1,000
-$1,500
Number of Garden Hoses Sold
350
400