A 4040 inch wire is to be cut. one piece is to be bent into the shape of a square, whereas the other piece is to be bent into the shape of a rectangle whose length is twice the width. find the width of the rectangle that will minimize the total area. what is the width of the rectangle that will minimize the total area?

Question

Alonzo Delacruz

Mathematics

9 July, 20:50

A 4040 inch wire is to be cut. one piece is to be bent into the shape of a square, whereas the other piece is to be bent into the shape of a rectangle whose length is twice the width. find the width of the rectangle that will minimize the total area. what is the width of the rectangle that will minimize the total area?

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Makai Espinoza · Answer 1

Let the square have sides s and the rectangle have sides r and 2r.

perimeters: 40 = 4s + 2 (r + 2r) = 4s + 6r → s = (1/4) (40 - 6r) = 10 - 1.5r areas: A = s² + 2r² = (10 - 1.5r) ² + 2r² = 100 - 30r + 4.25r²

Then dA/dr = 0 = 9.5r - 30 r = 30/9.5 = 3.16 in ← width of rectangle (length is twice the width) d²A/dr² = 9.5 > 0, so it is a minimum.