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# Profit, Revenue and Cost Problems for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these concepts can be found at: Applications of Derivatives Practice Problems for AP Calculus

### Summary of Formulas

1. P = RC: Profit = Revenue – Cost
2. R =xp: Revenue = (Units Sold)(Price Per Unit)
3. = Marginal Revenue ≈ Revenue from selling one more unit
4. = Marginal Profit ≈ Profit from selling one more unit
5. = Marginal Cost ≈ Cost of producing one more unit

Example 1

Given the cost function C(x) = 100 + 8x + 0.1x2,

1. find the marginal cost when x = 50; and
2. find the marginal profit at x = 50, if the price per unit is \$20.

Solution:

1. Marginal cost is C(x). Enter d(100 + 8x + 0.1x2, x)|x = 50 and obtain \$18.
2. Marginal profit is P '(x)
3. P = RC

P = 20x – (100 + 8x + 0.1x2). Enter d(20x – (100 + 8x + 0.1x^ 2, x)|x = 50 and obtain 2.

Example 2

Given the cost function C(x) = 500 + 3x + 0.01 x2 and the demand function (the price function) p(x) = 10, find the number of units produced in order to have maximum profit.

Solution:

Step 1:  Write an equation.

Profit = Revenue – Cost
P = RC
Revenue = (Units Sold)(Price Per Unit)
R =xp(x) = x(10) = 10x
P = 10x – (500 + 3x + 0.01x2)

Step 2:  :Differentiate.

Enter d(10x – (500 + 3x + 0.01x^2, x) and obtain 7 – 0.02x.

Step 3:  Find critical numbers.

Set 7 – 0.02x = 0 x = 350.
Critical number is x = 350.

Step 4:  Apply Second Derivative Test.

Enter d(10x – (500 + 3x + 0.01x^2), x, 2)|x = 350 and obtain –0.02.
Since x = 350 is the only relative maximum, it is the absolute maximum.

Step 5:  Write a Solution.

Thus, producing 350 units will lead to maximum profit.

Practice problems for these concepts can be found at: Applications of Derivatives Practice Problems for AP Calculus

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