Applications of Differentiation

Question

The revenue from selling q items is R(q) = 550q-q^2 , and the total cost is C(q) = 175+6q. Write a function that gives the total profit earned, and find the quantity which maximizes the profit.

Profit  pi(q)= ?

Quantity maximizing profit  q= 272

Answer 1

Profit is defined as -> P = R - C (if you earn $100 selling cookies, and you spend $25 making them, you have 75$ in profit).

So P = ( 550q-q^2 ) - ( 175 + 6q ) = 544q - 175 - q^2

P(q) = 544q - 175 - q^2

To maximize profit, you should take the derivative and set it equal to zero.

dP/dx = 544-2q = 0 --> q = 544/2 = 272

You can prove this is a maximum by using the second derivative test if need be.