Markus Petty

2022-07-20

You are designing a rectangular poster to contain 50 in^2 of printing with a 4-in. margin at the top and bottom and a 2-in margin at each side. What overall dimensions will minimize the amount of paper used?

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Let the paper size be x inches in length and y inches in width.
the length of the printed space would be x-8 inches and width would be y-4 inches. Print area would thus be (x-8)(y-4)= 50.
From this $y=4+\frac{50}{x-8}=\frac{4x+18}{x-8}$.
Also from the same equation on simplifying, it is xy-8y-4x+32=50.
Since the area of the paper of size x inches by y inches is xy, let it be denoted as A. Thus
A-8y-4x=18 Or A= 8y+4x+18
$A=\frac{32x+144}{x-8}+4x+18$. For minimum paper size $\frac{dA}{dx}$ must be =0, hence,
$\frac{dA}{dx}=\frac{32\left(x-8\right)-\left(32x+144\right)}{{\left(x-8\right)}^{2}}+4=0$
$\frac{-400}{{\left(x-8\right)}^{2}}+4=0$
${\left(x-8\right)}^{2}=100$
x-8=10
x=18, hence y=$4+\frac{50}{x-8}$=9
Dimension of minimum paper size would be 18 inches by 9 inches.

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