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A farmer has 100 yards of fencing to form two identical rectangular pens and a third pen that is twice as long as the other two...Asked by R L
A farmer has 100 yards of fencing to form two identical rectangular pens
and a third pen that is twice as long at the other two pens, as shown in the
diagram at right. All three pens have the same width, x. What value of y produces a maximum total fenced area?
and a third pen that is twice as long at the other two pens, as shown in the
diagram at right. All three pens have the same width, x. What value of y produces a maximum total fenced area?
Answers
Answered by
oobleck
You don't say how the pens are configured. I will assume that the long pen is adjacent to the other two, and they share a long side. So, if the pens have dimensions x*y, x*y and x*2y then the fencing needed is
5x+4y = 100
the area is
A = 2xy + x*2y = 4xy = 4x(100-5x)/4 = x(100-5x)
dA/dx = 100-10x
max area when x=10, so y = 50/4
As with all these problems, maximum area is achieved when the fencing is divided equally among lengths and widths.
5x+4y = 100
the area is
A = 2xy + x*2y = 4xy = 4x(100-5x)/4 = x(100-5x)
dA/dx = 100-10x
max area when x=10, so y = 50/4
As with all these problems, maximum area is achieved when the fencing is divided equally among lengths and widths.
Answered by
Anonymous
Oobleck's solution is wrong. The amount of fencing needed would actually be 5x + 6y = 100. His method to solve is still correct, though.
5x + 6y = 100
x = -(6/5)y+20
A = 2xy
A = 2y[-(6/5)y+20] = -12y^2/5 + 40y
A' = -24y/5 +40
24y = 200
y = 25/3
5x + 6y = 100
x = -(6/5)y+20
A = 2xy
A = 2y[-(6/5)y+20] = -12y^2/5 + 40y
A' = -24y/5 +40
24y = 200
y = 25/3
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