## Calculus: Early Transcendentals 8th Edition

The dimensions of the poster with the largest printed area are $~~10.95~in \times 16.44~in$
Let $w$ be the width of the poster. Let $h$ be the height of the poster. We can use the area of the poster to find an expression for the height $h$: $wh = 180$ $h = \frac{180}{w}$ We can find an expression for the area of the printed area: $A = (w-2)(h-3)$ $A = (w-2)(\frac{180}{w}-3)$ We can find $\frac{dA}{dw}$: $\frac{dA}{dw} = (\frac{180}{w}-3)+(w-2)(-\frac{180}{w^2}) = 0$ $180~w-3w^2+(w-2)(-180) = 0$ $180~w-3w^2-180~w+360 = 0$ $-3w^2+360 = 0$ $w^2-120 = 0$ $w^2 = 120$ $w = 10.95$ We can find $h$: $h = \frac{180}{w}$ $h = \frac{180}{10.95}$ $h = 16.44$ The dimensions of the poster with the largest printed area are $~~10.95~in \times 16.44~in$