# What dimensions should you make a rectangular poster of total area 72 square inches in order to maximize the printed area if it is required that the top and bottom margins be 2 inches and the side be 1 inch?

Nov 4, 2015

Let's call the width of the poster $x$ and the height $y$.

#### Explanation:

It's fairly easy to see that, because the surface area is
$x \cdot y = 72 \to y = \frac{72}{x}$, but we'll do that later.
Printable in $x$-direction: $x - 2 \cdot 1 = x - 2$
Printable in $y$-direction: $y - 2 \cdot 2 = y - 4$
Total printable area: $A = \left(x - 2\right) \cdot \left(y - 4\right) = x y - 4 x - 2 y + 8$

It's now time to substitute the $y$'s:
$A = x \cdot \left(\frac{72}{x}\right) - 4 x - 2 \cdot \left(\frac{72}{x}\right) + 8$
$A = \frac{72 \cancel{x}}{\cancel{x}} - 4 x - \frac{2 \cdot 72}{x} + 8 = 80 - 4 x - \frac{144}{x}$

For the optimum we need to set the deravative of $A$ to $0$

$A = 80 - 4 x - 144 {x}^{-} 1 \to A ' = 0 - 4 - \left(- 1\right) 144 {x}^{-} 2 \to$
Re-arrange and set to $0$
$144 {x}^{-} 2 - 4 = 0 \mathmr{and} \frac{144}{x} ^ 2 = 4 \to 4 {x}^{2} = 144 \to$
${x}^{2} = \frac{144}{4} = 36 \to x = 6$ (we only take the positive answer)
This is the width. The height is:
$y = \frac{72}{x} = \frac{72}{6} = 12$

Answer : Width is 6", height is 12".

Check :
Total area printed is $\left(6 - 2 \cdot 1\right) \left(12 - 2 \cdot 2\right) = 4 \cdot 8 = 32$ sq.in
Graph of the area-function $A = 80 - 4 x - \frac{144}{x}$
graph{80-4x-144/x [-36, 112.2, -11.25, 62.75]}