let the piece of paper be x inches wide and y inches long
area of printed part = (x-4)(y-8)
= xy - 8x - 4y + 32
but that equals 72
xy - 8x - 4y + 32 = 72
y(x-4) = 8x + 40
y = (8x+40)/(x-4) , clearly x > 4
amount of paper used = A
A = xy
= x(8x+40)/x-4)
= (8x^2 + 40x)/(x-4)
dA/dx = ((x-4)(16x+40) - (8x^2 + 40x)(1) )/(x-4)^2
= 0 for a min of A
(16x^2 - 24x - 160 - 8x^2 - 40x)/(x-4)^2 = 0
8x^2 - 64x - 160 = 0
x^2 - 8x - 20 = 0
(x - 10)(x + 2) = 0
x = 10 or x = -2 , the latter is not possible
x = 10 , then y = (8(10) + 40)/(10-4) = 20
The sheet of paper should be 10 inches wide and 20 inches long
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