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?

Jan 19, 2016

18" x 9"

Explanation:

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 $y = 4 + \frac{50}{x - 8} = \frac{4 x + 18}{x - 8}$.

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

$A = \frac{32 x + 144}{x - 8} + 4 x + 18$. For minimum paper size $\frac{\mathrm{dA}}{\mathrm{dx}}$ must be =0, hence,

$\frac{\mathrm{dA}}{\mathrm{dx}} = \frac{32 \left(x - 8\right) - \left(32 x + 144\right)}{x - 8} ^ 2 + 4 = 0$

$\frac{- 400}{x - 8} ^ 2 + 4 = 0$

${\left(x - 8\right)}^{2} = 100$

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.