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