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

### Summary of Formulas

*P*=*R*–*C*: Profit = Revenue – Cost*R*=*xp*: Revenue = (Units Sold)(Price Per Unit)- = Marginal Revenue ≈ Revenue from selling one more unit
- = Marginal Profit ≈ Profit from selling one more unit
- = Marginal Cost ≈ Cost of producing one more unit

**Example 1**

Given the cost function *C*(*x*) = 100 + 8*x* + 0.1*x*^{2},

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

Solution:

- Marginal cost is
*C*(*x*). Enter*d*(100 + 8*x*+ 0.1*x*^{2},*x*)|*x*= 50 and obtain $18. - Marginal profit is
*P*'(*x*)

*P* = *R* – *C*

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

**Example 2**

Given the cost function *C*(*x*) = 500 + 3*x* + 0.01 *x*^{2} 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*=

*R*–

*C*

- Revenue = (Units Sold)(Price Per Unit)

*R*=

*x*

*p*(

*x*) =

*x*(10) = 10

*x*

*P*= 10

*x*– (500 + 3

*x*+ 0.01

*x*

^{2})

Step 2: :Differentiate.

- Enter

*d*(10

*x*– (500 + 3

*x*+ 0.01

*x*^2,

*x*) and obtain 7 – 0.02

*x*.

Step 3: Find critical numbers.

- Set 7 – 0.02

*x*= 0

*x*= 350.

- Critical number is

*x*= 350.

Step 4: Apply Second Derivative Test.

- Enter

*d*(10

*x*– (500 + 3

*x*+ 0.01

*x*^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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