Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 32

Suppose that a poster with 1-inch margins as the bottom and sides and a 2-inch margin at the top have an area of 180in$^2$. What dimensions will give the largest printed area?







$ A = xy = 180;\\
\displaystyle y = \frac{180}{x}$


$A_1 = (x-2)(y-3)$

Substituting the value of $y$, we have

$
\begin{equation}
\begin{aligned}
A_1 &= (x - 2) \left( \frac{180}{x} - 3\right)\\
\\
A_1 &= 180 - 3x - \frac{360}{x} + 6
\end{aligned}
\end{equation}
$


Taking the derivative,

$
\begin{equation}
\begin{aligned}
A_1' &= -3 + \frac{360}{x^2}\\
\end{aligned}
\end{equation}
$


when $A_1' = 0,$

$
\begin{equation}
\begin{aligned}
0 & = - 3 + \frac{360}{x^2}\\
\\
\frac{360}{x^2} &= 3\\
\\
x^2 &= \frac{360}{3}\\
\\
x &= 2\sqrt{30}\text{ inches}
\end{aligned}
\end{equation}
$

when $x = 2\sqrt{30}\text{ inches,}$

$
\begin{equation}
\begin{aligned}
y &= \frac{180}{x} = \frac{180}{2\sqrt{30}} = 3 \sqrt{30}\text{ inches}
\end{aligned}
\end{equation}
$


Therefore, the dimensions that will give the largest printed area is $2 \sqrt{30}$in by $3\sqrt{30}$in