Find the dimensions of the rectangle with area 120 square inches and smallest possible perimeter

1 answer:

8 0

Here, A = L*W and P=2L+2W. Maximize P. Subst. L = A/W for L in the 2nd equation:

P=2(A/W) + 2W. This is to be minimized.

dP/dW = 2[-A/(W^2)) + 2W. But A = 120 sq in:

dP/dW = 2[-120/(W^2)) + 2W set this equal to 0 and solve for W.
Once you have W, find L: L=120/W

Let 2[-120/(W^2)) + W] = 0. Then [-120/(W^2)) + W] = 0

Multiplying all terms by W^2 gives us -120 + W^3 = 0.

w^3 = 120 cubic inches. Find the cube root to find W.

W = cube root of 120 = 4.93 inches. L = 120/W, or L = 120/4.93 inches

Summary: W = 4.93 inches and L = 24.33 inches (answer)

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