A 30 cm piece of wire is cut in two. One piece is bent into the shape of a rectangle with a length-to-width of 2:1. What are the lengths the two pieces if the sum of the areas of the square and rectangle is a minimum?

The other piece is bent into a square.

1 Answer
Ananda Dasgupta
Mar 29, 2018

#270/17# cm and #240/17# cm

Explanation:

Let the width of the rectangle be #x# cm. Then the length is #2x# cm, making the perimeter of the rectangle #6x# cm. This leaves #(30-6x) # cm for the square, whose side will have to be #3/2(5-x)# cm.

The total area now is #A" cm"^2# where

#A = 2x^2+(3/2(5-x))^2= 2x^2+9/4(5-x)^2#

so that

#(dA)/dx = 4x+9/4 times 2(5-x)times (-1) = 4x - 9/2(5-x)#

To maximize the area, we must set #(dA)/dx = 0#, leading to

# 4x - 9/2(5-x) = 0 implies (4+9/2)x = 9/2 times 5 #
#implies 17/2 x = 45/2 implies x = 45/17#

So, the length of the two pieces must be #45/17times 6 =270/17# cm, and #30-270/17 = 240/17# cm

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