Asked by Diane
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.
Answers
Answered by
Ms. Sue
Didn't you read my answer to your question?
http://www.jiskha.com/display.cgi?id=1380904270
http://www.jiskha.com/display.cgi?id=1380904270
Answered by
PsyDAG
The largest area would be closest to a square. Divide 1932 by 4.
Answered by
Diane
I tried that answer and it didn't work, the question wants the dimensions of the corral: length and width.
Answered by
Ms. Sue
Try PsyDAG's answer.
Answered by
Diane
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.
Answered by
Steve
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.
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.
Answered by
Diane
thank you Steve, that helped alot! appreciate your help.
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