5 Steps to a 5 AP Calculus BC 2019

Applications of Derivatives 159

Step 6: Using the Second Derivative Test, enterd(x∗(20− 2 x)∗(20− 2 x),x,2)|x=

10

3

and obtain−80. Thus,V

(

10

3

)

is a relative maximum. Since it is the only relative

maximum on the interval, it is the absolute maximum. (Note at the other endpoint

x=0,V(0)= 0 .)

Step 7: Therefore, the length of a side of the square to be cut isx=

10

3

.

TIP • The formula for the average value of a function f from x =a to x = b is
1
b−a

∫b

a

f(x)dx.

Business Problems

Summary of Formulas

1. P=R−C: Profit=Revenue−Cost


2. R=xp: Revenue=(Units Sold)(Price Per Unit)


3. C=

C

x

: Average Cost=

Total Cost

Units produced/Sold

4.

dR

dx

: Marginal Revenue≈Revenue from selling one more unit

5.

dP

dx

: Marginal Profit≈Profit from selling one more unit

6.

dC

dx

: Marginal Cost≈Cost of producing one more unit

Example 1

Given the cost functionC(x)= 100 + 8 x+ 0. 1 x^2 , (a) find the marginal cost whenx=50;

and (b) find the marginal profit atx=50, if the price per unit is $20.

Solution:

(a) Marginal cost isC′(x). Enterd(100+ 8 x+ 0. 1 x^2 ,x)|x=50 and obtain $18.

(b) Marginal profit isP′(x)

P=R−C

P= 20 x−(100+ 8 x+ 0. 1 x^2 ).Enterd(20x−(100+ 8 x+ 0. 1 x∧2,x)|x=50 and

obtain 2.

TIP • Carry all decimal places and round only at the final answer. Round to 3 decimal places
unless the question indicates otherwise.