Posted by **Anonymous** on Tuesday, October 25, 2011 at 7:26pm.

Show all work

The manufacturer of a CD player has found that the revenue R (in dollars) is R(p)= -4p^2+1280p, when the unit price is p dollars. If the manufacturer sets the price p to maximize revenue, what is the maximum revenue to the nearest whole dollar?

- introductory Algebra -
**drwls**, Tuesday, October 25, 2011 at 7:44pm
Find the value of p that maximizes R(p).

Since you don't know calculus yet, try completing the square.

R(p) = -4[p^2 -320 p + (160)^2] + 4*(160)^2

= 25,600 - 4(p - 160)^2

Maximum revenue occurs at p = 160.

## Answer This Question

## Related Questions

- calc2 - A manufacturer sells two products, one at a price of $3000 a unit and ...
- Calculus 2 - A manufacturer sells two products, one at a price of $3000 a unit ...
- Calculus 2 - A manufacturer sells two products, one at a price of $3000 a unit ...
- Math- Vector Calculus - A manufacturer sells two products, one at a price of $...
- College Math II - Solve the problem. If a manufacturer charges q dollars each ...
- algebra - The weekly cost of manufacturing x telephones per week is found by a ...
- calculus for business - The weekly cost of manufacturing x telephones per week ...
- College Algebra - Suppose that the price per unit in dollars of a cell phone ...
- algebra - Solve by using a system of equations. The number x of MP3 players a ...
- Further calculus - 1) A price p (in dollars) and demand x for a product are ...

More Related Questions