# Maximizing revenue

1. Nov 19, 2007

### Tekee

A company has 1000 widgets and will bbe able to sell all of them if the price is one dollar. The company will sell one less widget for each 10-cent increase in the price it charges. What price will maximize revenues, where revenue is the selling price times the quantity sold?

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I know that I have to make a system of equations, and then multiply the equations I get for price and quantity together, all under one variable. I take the derivative of that equation and find the maximum value...

However, I do not know how to start my equations which is, unfortunately, the hard part.

2. Nov 20, 2007

### HallsofIvy

Staff Emeritus
Well, you are told " company has 1000 widgets and will bbe able to sell all of them if the price is one dollar. The company will sell one less widget for each 10-cent increase in the price it charges. "
Okay, the number of widgets they can sell is 1000 minus the "drop off" If the price is x dollars then the increase in price (over one dollar) is x- 1. Since there are 10 10-cent increases in each dollar, that is 10(x-1) 10-cent increases in price so the number sold at price x is 1000-(10)(x-1).

10(x-1)= 10x- 10 so you can rewrite that as 1000- 10x+10= 1010- 10x. Of course, if the price is x dollars each, then the revenue is x times the number sold or just
x(1010- 10x)= 1010x- 10x^2. That's the revenue you want to maximize. Yes, you can do that by differentiating but I would consider "completing the square" to be more fundamental.