Ask Question
4 April, 21:10

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

+4
Answers (1)
  1. 5 April, 01:09
    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)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Find the dimensions of the rectangle with area 120 square inches and smallest possible perimeter ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers