Asked by Anonymous
A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100m^2. what are the dimensions of each pen that minimize the amount of fence that must be used?
Answers
Answered by
Steve
If each pen has width x and length y (against the barn),
the area is 4xy, and the fence used is 5x+8y
So, each pen has area 100.
p = 5x + 8(100/x) = 5x + 800/x
dp/dx = 5 - 800/x^2
dp/dx = 0 when x = √160 = 4√10
x = 4√10
y = 25/√10
the area is 4xy, and the fence used is 5x+8y
So, each pen has area 100.
p = 5x + 8(100/x) = 5x + 800/x
dp/dx = 5 - 800/x^2
dp/dx = 0 when x = √160 = 4√10
x = 4√10
y = 25/√10
Answered by
Steve
Now, if the barn is used as one wall of the pen, meaning only 3 sides have to be fenced, then
p = 5x+4y = 5x + 4(100/x) = 5x + 400/x
p' = 5 - 400/x^2
p' = 0 at x = 4√5
y = 25/√5
p = 5x+4y = 5x + 4(100/x) = 5x + 400/x
p' = 5 - 400/x^2
p' = 0 at x = 4√5
y = 25/√5
Answered by
Angel
The equation for the perimeter of the adjacent pens is incorrect. They are being built against the barn meaning that the farmer does not use fencing against the barn. The correct equation to model that situation would thus be 5x+4y.
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