Chapter 10 – 7a Marginal Analysis
In economics, the word “marginal” refers to an instantaneous rate of change (i.e., a derivative). Marginal cost is
the instantaneous change in the total cost relative to the number of units produced. If C(x) is the total cost to
produce x items, then C’(x) is the marginal cost. The marginal cost at a given value of x is an estimate of how much
it will cost to produce the next item (item x+1). This estimate is the differential of y (dy) for dx = 1. The actual cost
to product the next item is C(x+1)-C(x). This is the same as the increment in y (Δy) for Δx = 1.
𝐶 ′ (𝑥) ≈ 𝐶(𝑥 + 1) − 𝐶(𝑥) since 𝑑𝑦 ≈ ∆𝑦
Similarly, marginal revenue is the change in the total revenue relative to the number of items sold and marginal
profit is the change in the total profit relative to the number of items sold. The marginal revenue at x is
approximately equal to the revenue generated by selling the next item. The marginal profit is approximately equal
to the profit generated by selling the next item. If R(x) is the total revenue function then
𝑅′ (𝑥) ≈ 𝑅(𝑥 + 1) − 𝑅(𝑥)
𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) and 𝑃′ (𝑥) = 𝑅′ (𝑥) − 𝐶 ′ (𝑥) ≈ 𝑃(𝑥 + 1) − 𝑃(𝑥)
Example 1. The total cost (in dollars) of producing x food processors is
𝐶(𝑥) = 2000 + 50𝑥 − 0.5𝑥 2
A) Find the exact cost of producing the 21st food processor.
𝐶(21) − 𝐶(20) = $2,829.50 − $2,800.00 = $29.50
B) Use the marginal cost to estimate the cost of producing the 21st food
processor.
Total Cost of Production
$3,500
$3,000
$2,500
$2,000
$1,500
$1,000
$500
$0
0
5
10
15
20
25
30
Number of Food Processors
𝐶 ′ (𝑥) = 50 − 𝑥
𝐶 ′ (20) = 50 − 20 = $30.00
Example 2. The total cost (in dollars) of manufacturing x auto body frames is
$1,500
𝐶(𝑥) = 60,000 + 300𝑥
A) Find the average cost per frame if 500 frames are produced.
𝐶̅ (𝑥) =
𝐶(𝑥) 60,000 + 300𝑥 60,000
60,000
=
=
+ 300 ⟹ 𝐶̅ (500) =
+ 300
𝑥
𝑥
𝑥
500
= $420 per frame
B) Find the marginal average cost per frame at a production level of 500 frames and
interpret the results.
𝐶̅ ′ (𝑥) = −
60,000
𝑥2
⟹ 𝐶̅ ′ (500) = −
Average Cost of Production
60,000
5002
$1,000
$500
$0
0
500
1000
Number of Auto Body Frames
= −0.24. The average cost is decreasing at the rate of 24¢ a frame.
C) Use the results from A and B to approximate the average cost per frame if 501 frames are produced. Compare
your estimate to the actual average cost.
The approximate average cost is 420.00 – 0.24 = $419.76. The actual value is 𝐶̅ (501) = $419.7605.
Example 3. The total profit (in dollars) from the sale of x skateboards is
𝑃(𝑥) = 30𝑥 − 0.3𝑥 2 − 250
0 ≤ 𝑥 ≤ 100
A) Find the approximate profit from the sale of the 25th skateboard and compare it
to the actual profit.
Approximate: 𝑃′ (𝑥) = 30 − 0.6𝑥 ⟹ 𝑃′ (24) = $15.60
Actual: 𝑃(25) − 𝑃(24) = 312.5 − 297.2 = $15.30
Total Profit
$500
$300
$100
-$100 0
-$300
20
40
60
80
100
Number of Skateboards Sold
B) Find the approximate profit from the sale of the 90th skateboard and compare it to the actual profit.
Approximate: 𝑃′ (89) = −$23.40
Actual: 𝑃(90) − 𝑃(89) = 20 − 43.7 = −$23.70
Example 4. The total profit (in dollars) from the sale of x video DVDs is
𝑃(𝑥) = 5𝑥 − 0.005𝑥 2 − 450
Total Profit
0 ≤ 𝑥 ≤ 1,000
A) Evaluate the marginal profit at x = 450 and interpret the results.
𝑃′ (𝑥) = 5 − 0.01𝑥 ⟹ 𝑃′ (450) = $0.50. The total profit is increasing at the rate of
50¢ per DVD
B) Evaluate the marginal profit at x = 750 and interpret the results.
𝑃′ (750) = −$2.50. The total profit is decreasing at the rate of $2.50 per DVD.
$800
$600
$400
$200
$0
-$200 0
-$400
-$600
200
400
600
Number of DVDs
800
1000
Example 5. The demand (x units) for a clock radio depends on the price (p, in
dollars) as given by
Demand for Clock Radios
Demand
4000
𝑥 = 4000 − 40𝑝
3000
2000
1000
0
$0
$20
$40
$60
$80
$100
Unit Price
A) Find the price as a function of demand. What is the domain of this
function?
Solving the demand function for price: 𝑝 = 100 −
𝑥
40
0 ≤ 𝑥 ≤ 4000
Unit Price
$100
$80
$60
$40
$20
$0
0
1000
2000
3000
4000
Number of Clock Radios
B) Find the revenue, R(x), from the sale of x clock radios. What is the domain
of this function?
Total Revenue
$100,000
𝑅(𝑥) = 𝑥𝑝(𝑥) = 100𝑥 −
𝑥2
40
$80,000
0 ≤ 𝑥 ≤ 4000
$60,000
$40,000
$20,000
$0
0
1000
2000
3000
4000
Number of Clock Radios Sold
C) Find the marginal revenue at a production level of 1,600 clock radios and
interpret the results.
𝑅′ (𝑥) = 100 −
𝑥
20
⟹ 𝑅′ (1600) = $20. Total revenue is increasing at the
rate of $20 per clock radio.
Marginal Revenue
$100
$50
$0
-$50
D) Find the marginal revenue at a production level of 2,500 clock radios and
interpret the results.
𝑅′ (2500) = −$25. Total revenue is decreasing at the rate of $25 per clock radio.
-$100
0
1000
2000
3000
Number of Clock Radios
4000