#### Answer

The dimensions of the poster with the largest printed area are $~~10.95~in \times 16.44~in$

#### Work Step by Step

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 $~~10.95~in \times 16.44~in$