Question

A poster must have 32 square inches of printed matter with margins of 4 inches at the top and bottom, and 2 inches at each side. Find the dimensions of the whole poster if its area is maximum.

Answer 1

$xy=32\to y=\frac{32}{x}.$

$A=(x+4)(y+8)=xy+8x+4y+32=32+8x+\frac{128}{x}+32=8x+\frac{128}{x}+64.$

$\frac{dA}{dx}=0 \to 8-\frac{128}{x^2}=0 \to x^2=16 \to x=4.$

$y=\frac{32}{4}=8.$

The dimensions of the whole poster if its area is maximum are 4 in by 8 in.