Answer to Question #250627 in Calculus for dates s

Question

Johnny is designing a rectangular poster to contain 48inches squared of printing with a 4-in margin at the top and bottom and a 3-in margin at each side. What overall dimensions will minimize the amount of paper​ used?

Answer 1

Let the dimensions of the poster be x by y inches.

$xy=48\to y=\frac{48}{x}.$

$A=(x+6*2)(y+8*2)=xy+16x+12y+192=16x+\frac{576}{x}+240.$

$\frac{dA}{dx}=0\to16-\frac{576}{x^2}=0\to x^2=36.$

$x=\sqrt{36}=6.$

$y=\frac{48}{6}=8.$

$x+6*2=6+12=18.$

$y+8*2=8+16=24.$

The overall dimensions of the paper: 18 by 24 inches.