SOLUTION: A piece of wire 20 ft long is cut into two pieces. One piece is bent into the shape of a circle and the other one into the shape of square. How should the wire be cut so that a.

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Question 1176607: A piece of wire 20 ft long is cut into two pieces. One piece is bent into the shape of a circle and the other one into
the shape of square. How should the wire be cut so that
a. the combined area of the two figures is as small as possible?
b. the combined area of the two figures is as large as possible?
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


Let x be the length of wire allocated for the circle

Then  20-x is allocated for the square



A_circle = pi%2Ar%5E2 where +r=x%2F%282%2Api%29+ -->  A_circle = x%5E2%2F%284%2Api%29+



A_square = +%28%2820-x%29%2F4%29%5E2+ = +%28400-40x%2Bx%5E2%29%2F16+



A = A_total = x%5E2%2F%284%2Api%29+ + +%28400-40x%2Bx%5E2%29%2F16+   (*)



dA%2Fdx = x%2F%282%2Api%29+ + +%28-40%2B2x%29%2F16+



[ Note d%5E2A%2Fdx%5E2+ = +1%2F%282%2Api%29+%2B+2%2F16+ ---> concave up --> setting dA/dx=0 will find a MINIMUM ]



+dA%2Fdx+ = x%2F%282%2Api%29+ + +%28-40%2B2x%29%2F16+ = 0

+x+=+80%2Api%2F%284%2Api%2B16%29+ or approx. +highlight%288.798%29+ft





A_min = 8.798%5E2+%2F+%284%2Api%29+ + +%28%2820-8.798%29%2F4%29%5E2+

      = +6.1597+ +ft%5E2+ + +7.843+ft%5E2+

      = +14.003+ft%5E2+





For A_max, you must check endpoints (i.e. x=0ft and x=20ft).  Here you will find that if you allocate all of the wire for the circle then the area will be maximized.



This graph of A_total (eq (*)) illustrates this: