A4040 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?

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.

No because if you do the higher number minus the lower number you will get a positive. if you do it the other way around you will get a negative number