SOLUTION: A piece of wire 10 m long is to be cut into two pieces. One piece is bent into a square, and the other into a circle. How can the wire be cut so that the area enclosed by the two s
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-> SOLUTION: A piece of wire 10 m long is to be cut into two pieces. One piece is bent into a square, and the other into a circle. How can the wire be cut so that the area enclosed by the two s
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Question 1152181: A piece of wire 10 m long is to be cut into two pieces. One piece is bent into a square, and the other into a circle. How can the wire be cut so that the area enclosed by the two shapes is minimized? Give your answer to the nearest tenth. Answer by math_helper(2461) (Show Source):
Given L=10m, find c such that a circle made from the material 0..c and a
square made from the material c..L have minimum area.
A_circle =
A_square =
A_total (= A) = A_circle + A_square =
Take derivative of A wrt c:
dA/dc = =
[ Note that so the curve is concave up, thus the critical point we are about to find is a local minimum]
Set first deriv. to zero & solve for c: c = 4.39901m (or to one decimal place)
I have separately verified that this value of c minimizes the total area. You should also verify it to convince yourself. Also a good idea is to plug in the extreme values of c (0, 10) to gain an understanding of the possible areas. Here is a graph of A(c) to help you visualize:
