Question

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?

Answer 1

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+ 50/(x-8) = (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= (32x+144)/(x-8) +4x +18`. For minimum paper size `(dA)/dx` must be =0, hence,

`(dA)/dx= (32(x-8) - (32x+144))/(x-8)^2 +4 =0`

`(-400)/(x-8)^2 +4=0`

`(x-8)^2 =100`

x-8=10
x=18, hence y= `4+ 50/(x-8)` =9

Dimension of minimum paper size would be 18 inches by 9 inches.