Posted by jemma on Friday, November 8, 2013 at 4:01am.
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?

math  Steve, Friday, November 8, 2013 at 5:15am
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
(x2)/4 by (x2)/4 + 1 or
(x2)/4 by (x+2)/4
and
(y2)/4 by (y+2)/4
But, y = 36x, so the rectangles are
(x2)/4 by (x+2)/4
(34x)/4 by (38x)/4
and the sum of their areas is
(x2)(x+2)/16 + (34x)(38x)/16
= (x^236x+644)/8
= 1/8 [(x18)^2 + 320]
This is a parabola with vertex at (18,40), where the area is a minimum.

or  math  Reiny, Friday, November 8, 2013 at 7:55am
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 = 8x
area = A = x(x+1) + y(y+1)
= x^2 + x + y^2 + y
= x^2 + x + 6416x+x^2 + 8x
= 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
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