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December 21, 2014

December 21, 2014

Posted by **Cole** on Friday, October 5, 2012 at 8:00am.

- Precalculus -
**Reiny**, Friday, October 5, 2012 at 8:28amDid you make a sketch ?

Even though you didn't say, I will assume that the corrals are joined with common lengths.

let the width of the entire corral (parallel to the barn) be y ft

let each width of corrals be x ft

So we have y + 5x = 200

y = 200 - 5x

Area = xy = x(200-5x) or 200x - 5x^2

The equation will vary depending on your definition of the variables. Mine avoids unnecessary fractions .

BTW, after you maximize the area function, the dimensions of each corral will be 20 by 25, for a maximum area of 2000 for the combined area

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