Thread: Optimization Problem

Hi,
I have math homework so I could do all the questions but two problems... I'll post one of them because the other one is pretty the same.

1) The TruTime Watch Company has been selling 1200 watches per week at $18 each. They are planning a price increase. A survey indicates that for every dollar increase in price there will be a drop of 40 sales per week. What should the selling price be in order to maximize revenue?

Thanks

Originally Posted by dgolverk

1) The TruTime Watch Company has been selling 1200 watches per week at $18 each. They are planning a price increase. A survey indicates that for every dollar increase in price there will be a drop of 40 sales per week. What should the selling price be in order to maximize revenue?

Let x be the increase in price, and f(x) the weekly revenue.

f(x)="number of watches sold per week" times "price per watch" =(1200-40x)*(18+x)

Calculate the derivative f´(x) and solve the equation f´(x)=0. The solution of that equation, plus 18, is your answer. The result I get is $24.

You know the solution of f´(x)=0 is a maximum for the function f (rather than a minimum) by calculating the second derivative f"(x). It's negative, and that means it's a maximum.