College Algebra

Question

P(x)= -x^2+90x-300 is a profit function where x is the number of items sold, and P(x) is the profit from that sale. Find the maximum profit and the number of items that must be sold to reach that profit.

Answer 1

take the derivative of P(x), set it =0, solve for x

P'(x)= 2x +90 = 0

x = 90/2 = 45 items sold maximizes profit

max profit =P(45) = (45)^2 +90(45 -300

= 2025 +4050 -300

= 6075 -300

= $5,775

or without calculus, just use algebra

and rewrite the P(x) function in vertex form, by completing the square

= (x-45)^2 + 5775 which has vertex = (45, 5775) = (profit maximizing output, max profit)