Helppppp pleaseeeee!!!!

Question

The manger of a home Electronics store is considering re-pricing a new model of digital camera at the current price of $680 the store sells about 70 cameras each month sales data from other stores indicate that for each $20 decrease in price about five more cameras per month would be sold how much should the manager charge for a Camera to maximize monthly revenue? What is the maximum revenue?

Answer 1

If you plot the Total Money vs the Number of Cameras Sold, you will get the following equation:

Money = -4*Camera² + 960*Camera

When you differentiate dM/dC = -8*Camera + 960 (Set this to 0 to find the maximum)

Setting this = 0, 8*Camera = 960 & you find that the Number of Cameras for a maximum = 120 Cameras.

The highest total money for 120 Cameras is -4*(120)² + 960*Cameras = $57,600