Posted by **Diane** on Friday, October 4, 2013 at 12:54pm.

A rancher wishes to enclose a rectangular partitioned corral with 1932 feet of fencing. What dimensions of the corral would enclose the largest possible area? Find the maxium area.

- College Algebra -
**Ms. Sue**, Friday, October 4, 2013 at 12:57pm
Didn't you read my answer to your question?

http://www.jiskha.com/display.cgi?id=1380904270

- College Algebra -
**PsyDAG**, Friday, October 4, 2013 at 12:58pm
The largest area would be closest to a square. Divide 1932 by 4.

- College Algebra -
**Diane**, Friday, October 4, 2013 at 1:00pm
I tried that answer and it didn't work, the question wants the dimensions of the corral: length and width.

- College Algebra -
**Ms. Sue**, Friday, October 4, 2013 at 1:01pm
Try PsyDAG's answer.

- College Algebra -
**Diane**, Friday, October 4, 2013 at 1:04pm
Tried that too and still didn't work. the picture has a rectangular fence with a fence going across the middle from longest side to the opposite side.

- College Algebra -
**Steve**, Friday, October 4, 2013 at 3:27pm
well, if you want a correct answer, you need to pose the correct question. If the short side is x and the long side is y, then

2y+3x = 1932, so y = (1932-3x)/2

area = xy = x(1932-3x)/2 = (1932x-3x^2)/2

This is a parabola, with vertex at x=322.

You can probably take it from there. Note that just as PsyDiag suggested, the fencing is evenly divided between length and width.

- College Algebra -
**Diane**, Friday, October 4, 2013 at 3:51pm
thank you Steve, that helped alot! appreciate your help.

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