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

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

anonymous · Answer

P= -s² + 120s – 2000
Thus the maximum profit is $1,600 when unit price is $60
Compare with profit is $1,500 at unit price $50
or loss $ 900 if the sale price is $10!