Asked by Henry
A farmer wishes to build a fence for 6 adjacent rectangular pens. If there is 600 feet of fencing available, what are the dimensions of each pen that maximizes total pen area?
The image looks like this:
box box box
box box box
also:
If the interior fencing is $3.00 per foot and the perimeter is $5.00 per foot, what are the pen dimensions that minimize cost?
The image looks like this:
box box box
box box box
also:
If the interior fencing is $3.00 per foot and the perimeter is $5.00 per foot, what are the pen dimensions that minimize cost?
Answers
Answered by
Henry
assume each pen encloses 200 square feet of area in the 2nd part sorry!
Answered by
Steve
part 2:
if each pen has width x and height y
xy=200
6x+8y=600
you have no room to vary the dimensions. So, I assume the 600 feet of fencing does not apply to part 2. Accordingly,
xy=200
cost c = 5(3x+4y)+3(3x+4y)
= 24x+32y
= 24x+32(200/x)
dc/dx = 24 - 6400/x^2
= 8(3x^2-800)/x^2
dc/dx = 0 when x = 20√(2/3)
so, the pens are 20√(2/3) by 10√(3/2)
if each pen has width x and height y
xy=200
6x+8y=600
you have no room to vary the dimensions. So, I assume the 600 feet of fencing does not apply to part 2. Accordingly,
xy=200
cost c = 5(3x+4y)+3(3x+4y)
= 24x+32y
= 24x+32(200/x)
dc/dx = 24 - 6400/x^2
= 8(3x^2-800)/x^2
dc/dx = 0 when x = 20√(2/3)
so, the pens are 20√(2/3) by 10√(3/2)
Answered by
Chris
25'x25'
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