A 36 inch wire is to be cut One piece is to be bent into the

A 36 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? (Round to the nearest tenth as needed.)

Solution

let width of recatance be x
Then length of rectangle will be 2*x

Let the length of side of square be a

Total perimeter:
perimeter of rectangle + perimeter of square = 36
2(x+2x) + 4*a = 36
6x + 4*a = 36
a = (36-6x)/4
= 9 - 1.5*x

Total area,
A = Area of rectangle + area of square
= x*2x + (a)^2
= 2x^2 + (9-1.5x)^2
= 4.25*x^2 - 27*x + 81

A\'(x) = 8.5*x-27
put A\'(x) = 0
8.5*x= 27
x = 54/17 inch

This givies minimum area
Answer: 54/17 inch