Consider the following. P = −0.1s3 + 6s2 + 400.

Required:
a. Find the amount s of advertising (in thousands of dollars) that maximizes the profit P (in thousands of dollars).
b. Find the point of diminishing returns.

A) s = $40 (in thousands of dollars)

B) point of diminishing returns is at;

(20, 2000) in thousands of dollars

Step-by-step explanation:

We are given the profit function as;

P = −0.1s³ + 6s² + 400

A) To maximize the profit, we need to find the first derivative and equate it to zero.

Thus;

dP/ds = -0.3s² + 12s

At dP/ds = 0, we have;

-0.3s² + 12s = 0

0.3s² = 12s

0.3s = 12

s = 12/0.3

s = $40 (in thousands of dollars)

B) To find the point of diminishing returns, we need to find the 2nd derivative of the given profit function and equate to zero.

Thus;

d²P/ds² = -0.6s + 12

At d²P/ds² = 0, we have;

-0.6s + 12 = 0

0.6s = 12

s = 12/0.6

s = 20

At s = 20,

P = −0.1(20)³ + 6(20)² + 400

P = -800 + 2400 + 400

P = 2000

Thus; point of diminishing returns is at;

(20, 2000) in thousands of dollars

4th option is correct xy=-10

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