# Question #5cf1a

May 3, 2017

The smallest area is $864 \setminus c {m}^{2}$ which occurs when the dimensions are $36 \setminus c m \times 24 \setminus c m$

#### Explanation:

Let us set up the following variables:

$\left\{\begin{matrix}x & \text{Width of poster (cm)" \\ y & "Height of poster (cm)" \\ A & "Area of poster ("cm^3")}\end{matrix}\right.$

Then the dimensions of the printed matter are:

$\left\{\begin{matrix}\text{Width" & =x-6-6 & =x-12 \\ "Height" & =y-4-4 & =y-8 \\ :. " Area} & = 384 & = \left(x - 12\right) \left(y - 8\right)\end{matrix}\right.$

So we have;

$384 = \left(x - 12\right) \left(y - 8\right)$
$\therefore y - 8 = \frac{384}{x - 12}$
$\therefore y = 8 + \frac{384}{x - 12}$
$\text{ } = \frac{8 \left(x - 12\right) + 384}{x - 12}$
$\text{ } = \frac{8 x - 96 + 384}{x - 12}$
$\text{ } = \frac{8 x + 288}{x - 12} \setminus \setminus \setminus \setminus \setminus \ldots . . \left(\star\right)$

And the total area of the poster is given by:

$A = x y$
$\setminus \setminus \setminus = \left(x\right) \left(\frac{8 x + 288}{x - 12}\right) \setminus \setminus \setminus$ (using ($\star$))
$\setminus \setminus \setminus = \frac{8 {x}^{2} + 288 x}{x - 12}$

We want to minimize (hopefully) by finding $\frac{\mathrm{dA}}{\mathrm{dx}}$, which we get by applying the product rule:

$\frac{\mathrm{dA}}{\mathrm{dx}} = \frac{\left(x - 12\right) \left(16 x + 288\right) - \left(1\right) \left(8 {x}^{2} + 288 x\right)}{x - 12} ^ 2$
$\text{ } = \frac{16 {x}^{2} + 288 x - 192 x + 3456 - 8 {x}^{2} - 288 x}{x - 12} ^ 2$
$\text{ } = \frac{8 {x}^{2} - 192 x + 3456}{x - 12} ^ 2$

At a min or max $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$

$\therefore \frac{8 {x}^{2} - 192 x + 3456}{x - 12} ^ 2 = 0$
$\therefore 8 {x}^{2} - 192 x - 3456 = 0$
$\therefore 8 \left({x}^{2} - 24 x - 432\right) = 0$
$\therefore {x}^{2} - 24 x - 432 = 0$
$\therefore \left(x + 12\right) \left(x - 36\right) = 0$

This equation leads to the two solutions:

$x = - 12$, or $x = 36$

Obviously $x > 0$, so we can eliminate $x = - 12$, leaving $x = 36$ as the only valid solution. With this value of $x$ we have:

$y = \frac{288 + 288}{36 - 12} \setminus \setminus \setminus \setminus \setminus$ (using ($\star$))
$\setminus \setminus = \frac{576}{24}$
$\setminus \setminus = 24$

With these dimensions we have:

$A = 36 \cdot 24$
$\setminus \setminus \setminus = 864$

We should check that this value leads to a minimum (rather than a maximum) area, which we confirm graphically:

graph{(8x^2+288x)/(x-12) [-2, 50, -2000, 2000]}