SOLUTION: A farmer has 3000 feet of wire to enclose a rectangular field. He plans to fence the entire area and then subdivide it by running a perpendicular fence across the middle. Find the
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Question 813329: A farmer has 3000 feet of wire to enclose a rectangular field. He plans to fence the entire area and then subdivide it by running a perpendicular fence across the middle. Find the dimensions of the field that would enclose the maximum area . What is the maximum area?
Can you please help me out? Thanks so much in advance:)
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
A = Lw
P = 2L + 2w = 3000
2L + 2w = 3000
2L = 3000 - 2w
L = 1500 - w
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A = Lw
A = (1500 - w)w
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A(w) = -w^2 + 1500w + 0
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the above quadratic function is represented by the quadratic equation in standard form, with a=-1, b=1500, and c=0
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to find the maximum area A(w), plug this:
-1 1500 0
into this: https://sooeet.com/math/quadratic-equation-solver.php
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Answer 1:
the maximum point of the above quadratic equation is: ( w= 750, A(w)= 562500 )
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so the maximum area that can be enclosed is: 562500 sq.ft
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now, to find the L and w dimensions of the maximum area, remember this equation from above:
A(w) = -w^2 + 1500w + 0
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we used that equation to maximize Area, but as a side effect we also got w for the max area, from the max point of the quadratic equation ( w= 750, A(w)= 562500 )
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Answer 2:
w for maximum area = 750
L for maximum area = 1500 - w = 1500 - 750 = 750
so the maximum area has dims: w=750, L=750 feet
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