Question

A cell phone manufacturer makes profits (P) depending on the sale price (s) of each phone. The function P= -s² + 120s – 2000 models the monthly profit for a flip phone from Cellular Heaven. What should the phone price be to make the maximum profit? What is the maximum profit possible?Steps to solve this application problem:
1. Find the vertex (x, y)= (x= sale price of phone, y= Profit on phone sales)
2. The x-coordinate is the price of the phone when the profit is a maximum
3. The y-coordinate is the maximum possible profit
Answer
a. price $10, max profit $600
b. price $50, max pro

Answer 1

ITS 6x

Answer 2

(100,6000)

Answer 3

these are choices

Answer 4

a. price $10, max profit $600
b. price $50, max profit $1200
c. price $6, max profit $100
d. price $60, max profit $1600

Answer 5

well,I like to use derivative for this one , It isn`t the simplest, but it`s fastest. so calculate the derivative and find its root: -2s+120=0 , s=60, we know that the derivatives root is where the function is in maximum. so the price should be 60 and the profit is f(60)=1600. choice d.