Question

You are designing a rectangular poster to contain 50 in2 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

Answer:

W = 7.53 inches

H = 30.14 inches

Explanation:

Printing area is:

$A_p=W_o*H_o=50 in^2$

Solving for $H_o$:

$H_o=50/W_o$

The total area of the paper is:

$A_t=(W_o+4)*(H_o+16)$   Replacing $H_o$:

$A_t=(W_o+4)*(50/W_o+16)$

If we derivate and equal zero to find the minimum:

$A_t'=(50/W_o+16)+(W_o+4)*(-50/W_o^2)$

$0=(50/W_o+16)+(W_o+4)*(-50/W_o^2)$

$0=16-200/W_o^2$

$W_o=\sqrt{200/16}$

$W_o=3.53-inch$   So,

$H_o=14.14-inch$

The final dimensions of the paper are:

W = 3.53 + 4=7.53-inch

H = 14.14 + 16= 30.14-inch