Question

the revenue from selling $q$ items is $r(q) = 575q - q^2$, and the total cost is $c(q) = 125 + 11q$.
write a function that gives the total profit earned, find its derivative, and find the quantity which maximizes the profit. (note: $p(q) = r(q) - c(q))$.
profit $p(q) = $
derivative $p(q) = $
quantity maximizing profit $q = $

Explanation:

Step1: Define profit function

$P(q) = R(q) - C(q) = (575q - q^2) - (125 + 11q)$
Simplify: $P(q) = 575q - q^2 - 125 - 11q = -q^2 + 564q - 125$

Step2: Compute profit derivative

$P'(q) = \frac{d}{dq}(-q^2 + 564q - 125)$
Apply power rule: $P'(q) = -2q + 564$

Step3: Find critical point

Set $P'(q) = 0$:
$-2q + 564 = 0$
Solve for $q$: $2q = 564 \implies q = \frac{564}{2} = 282$

Answer:

Profit $P(q) = -q^2 + 564q - 125$
Derivative $P'(q) = -2q + 564$
Quantity maximizing profit $q = 282$