Question

a wire is 36 m long is cut into two pieces , each piece is bent to form a rectangle which is 1 centimeter longer than its width. how long should each piece be to minimize the sum of the areas of the two rectangles?

Answer 1

seems like I saw this one go by the other day, but here it is again:

If the two pieces are x and y in length, then the rectangles have dimensions

(x-2)/4 by (x-2)/4 + 1 or
(x-2)/4 by (x+2)/4
and
(y-2)/4 by (y+2)/4

But, y = 36-x, so the rectangles are

(x-2)/4 by (x+2)/4
(34-x)/4 by (38-x)/4

and the sum of their areas is

(x-2)(x+2)/16 + (34-x)(38-x)/16
= (x^2-36x+644)/8
= 1/8 [(x-18)^2 + 320]

This is a parabola with vertex at (18,40), where the area is a minimum.

Answer 2

avoiding some of those fractions

let the sides of one rectangle be x and x+1
and the other, y and y+1
4x+2 + 4y+2 = 36
x+y=8
y = 8-x

area = A = x(x+1) + y(y+1)
= x^2 + x + y^2 + y
= x^2 + x + 64-16x+x^2 + 8-x
= 2x^2 - 16x + 72
Quadratic opening up, so a minimum
dA/dx = 4x - 16
= 0 for a min
x = 4
then y = 4

The two rectangles are the same size, both 4 by 5
so we need 2(4+5) or 18 m for each
for a min area of 4(5) + 4(5) or 40 m^2
for a min