Wednesday

January 28, 2015

January 28, 2015

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

**Answer this Question**

**Related Questions**

Precalculus - There are 4 rectangular corrals of identical dimensions along the ...

Precalculus - A barn has 150 feet of fencing and there are 3 rectangular corrals...

Precalculus - A barn has 150 feet of fencing and there are 3 rectangular corrals...

Precalculus - Combined area of 1680ft^2 for 4 identical rectangular corrals. ...

Precalculus - I have a diagram that has 4 rectangular corrals joined together ...

Math- Please Help Us! - Diagram has 4 rectangular corrals with a barn above it. ...

Math: pre calculus - a barn is 150 feet long and 75 feet wide. the owner has 240...

algebra - a rancher has 310 feet of fencing with which to enclose two ...

Precalculus - I usually know how to do these types of problems, but the second ...

math - a rectangular garden with an area of 2112 square feet is to be located ...