SOLUTION: A rectangular field is to be enclosed with a fence. One side of the field is against an existing fence, so that no fence is needed on that side. If material for the fence cost $2 p

Question 296318: A rectangular field is to be enclosed with a fence. One side of the field is against an existing fence, so that no fence is needed on that side. If material for the fence cost $2 per foot for the two ends and $4 per foot for the side parallel to the existing fence, find the dimensions of the field of the largest area that can be enclosed for 1,000.
Thanks in Advance
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A rectangular field is to be enclosed with a fence.
One side of the field is against an existing fence, so that no fence is needed on that side.
If material for the fence cost $2 per foot for the two ends and $4 per foot for
the side parallel to the existing fence, find the dimensions of the field of
the largest area that can be enclosed for 1,000.
:
Let L = length of the $2 sides
Let W = length of the $4 side
:
2(2L) + 4W = 1000
4L + 4W = 1000
Simplify, divide by 4
L + W = 250
L = (250-W)
:
Area:
A = L*W
Replace L with (250-W)
A = (250-W) * W
A = -W^2 = 250W
Find the axis of symmetry: a=-1; b=250, will give max area
W =
W = +125 ft, length of $4 side
then
L = 250-125 = 125 ft, length of the $2 sides
:
Max area: all three sides = 125 ft
:
:
Check 2(2*125) + 4(125) = $1000

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