a poster is to have a total area of 180 in² with a 1 - inch margins at the bottom and the sides and a 2…

Explanation:

Step1: Let the length be $x$ and width be $y$.

The area of the poster is $A = xy=180$, so $y=\frac{180}{x}$. The printed - area $P=(x - 2)(y - 3)$.

Step2: Substitute $y=\frac{180}{x}$ into the printed - area formula.

$P=(x - 2)(\frac{180}{x}-3)=180-3x-\frac{360}{x}+6=186-3x-\frac{360}{x}$, where $x>2$ and $y>3$.

Step3: Find the derivative of $P$ with respect to $x$.

$P^\prime=-3+\frac{360}{x^{2}}$.

Step4: Set the derivative equal to zero to find critical points.

$-3+\frac{360}{x^{2}} = 0$. $\frac{360}{x^{2}}=3$. $x^{2}=120$, so $x = \sqrt{120}=2\sqrt{30}\approx10.95$.

Step5: Find the second - derivative of $P$ to determine if it's a maximum.

$P^{\prime\prime}=-\frac{720}{x^{3}}$. When $x = 2\sqrt{30}$, $P^{\prime\prime}<0$, so it's a maximum.

Step6: Find the value of $y$.

Since $y=\frac{180}{x}$, when $x = 2\sqrt{30}$, $y=\frac{180}{2\sqrt{30}} = 3\sqrt{30}\approx16.43$.

Answer:

The poster with dimensions $x = 2\sqrt{30}\text{ inches}\approx10.95\text{ inches}$ and $y = 3\sqrt{30}\text{ inches}\approx16.43\text{ inches}$ gives the largest printed area.