Question

A poster is to have an area of $210{\mathrm{inch}}^2$ with $1\mathrm{inch}$ margins at the bottom and sides and a $2\mathrm{inch}$ margin at the top.

Find the exact dimensions that will give the largest printed area.

Answer 1

Finding the exact dimensions of the poster which will give the largest area:

Given, area is $210$

Margin at bottom and side is $1\mathrm{inch}$

Margin at top is $2\mathrm{inch}$

Let the length be $\mathrm{x}\mathrm{inch}$ and width be $\mathrm{y}\mathrm{inch}$

Expressing $\mathrm{y}$ in terms of $\mathrm{x}$

$$\begin{gathered} \mathrm{Area}=\mathrm{length}\times \mathrm{width} \\ \Rightarrow 210=\mathrm{xy} \\ \Rightarrow \mathrm{y}=\frac{210}{\mathrm{x}} \end{gathered}$$

The printed area will be the area excluding the margins.

Calculating the printed area:

The printed area can be written as,

$\begin{array}{rcl}A& =& \mathrm{length}\times \mathrm{width}\\ & =& \left(\mathrm{x}-3\right)\left(\mathrm{y}-2\right)\\ & =& \left(x-3\right)\left(\frac{210}{x}-2\right)\\ & =& 210-2x+6-\frac{630}{x}\end{array}$

Now, in order to find the maximum printed area, we will have to calculate the critical point.

Finding the critical point:

Differentiating the area with respect to $\mathrm{x}$ to find the critical point.

$\begin{array}{rcl}\frac{\mathrm{dA}}{\mathrm{dx}}& =& -2+\frac{630}{{\mathrm{x}}^2}\\ & \Rightarrow & 0=-2+\frac{630}{{\mathrm{x}}^2}\left[\frac{\mathrm{dA}}{\mathrm{dx}}=0\mathrm{for}\mathrm{maximizing}\mathrm{A}\right]\\ & \Rightarrow & \frac{630}{{\mathrm{x}}^2}=2\\ & \Rightarrow & {\mathrm{x}}^2=315\\ & \Rightarrow & \mathrm{x}=\sqrt{315}\\ & =& 3\sqrt{35}\\ \therefore \mathrm{y}& =& \frac{210}{\mathrm{x}}=\frac{210}{3\sqrt{35}}=\sqrt{\frac{70\times 70}{35}}=2\sqrt{35}\end{array}$

Hence, the dimensions are $3\sqrt{35}\mathrm{inch},2\sqrt{35}\mathrm{inch}$.