Quantity is profit

Question

Revenue is given by R(q) = 600q

 and cost is given by C(q) = 7000 + 3q².

 At what quantity is profit maximized?

q =

*** I got 100 and it was correct

What is the total profit at this production level?

$_____

*** I got 2000 and it was incorrect. Please help!

Answer 1

You got the first part correct:

Since Profit = Revenue - Cost, then P(q) = R(q) - C(q) = 600q - (7000 + 3q²) or

P(q) = -3q² + 600q - 7000

To find the maximum, take the derivative and set it equal to zero.

P'(q) = -6q + 600

-6q + 600 = 0

q = 100

But the profit when q = 100 comes from plugging 100 into P(q)

P(100) = -3(100)² + 600(100) - 7000 = $23,000