A rectangle is to have an area of 16 square inches. How do you find its dimensions so that the distance from one corner to the midpoint of a nonadjacent side is a minimum?

Jan 8, 2016

$2 \sqrt{2}$ $\text{in} \times 4 \sqrt{2}$ $\text{in}$

Explanation:

We can write the following equations:

$l w = 16$

Draw a diagram of the line cutting through the rectangle and use the Pythagorean Theorem to say that the length of the segment can be found through:

$f \left(l , w\right) = \sqrt{{l}^{2} + {\left(\frac{w}{2}\right)}^{2}}$

Using the area equation, we can make $f \left(l , w\right)$ into a single variable equation by substituting.

$l = \frac{16}{w}$

Thus,

$f \left(w\right) = \sqrt{{\left(\frac{16}{w}\right)}^{2} + {\left(\frac{w}{2}\right)}^{2}}$

Simplify:

$f \left(w\right) = \sqrt{\frac{256}{w} ^ 2 + {w}^{2} / 4} = \sqrt{\frac{1024 + {w}^{4}}{4 {w}^{2}}} = \frac{\sqrt{{w}^{4} + 1024}}{2 w}$

It should be noted that the domain of this function, or the values for which $w$ can exist, is $0 < w < \infty$.

To find the minimum value, find the derivative of $f \left(w\right)$ through the quotient rule (or product rule).

$f ' \left(w\right) = \frac{\frac{4 {w}^{3} \left(2 w\right)}{2 \left(\sqrt{{w}^{4} + 1024}\right)} - 2 \sqrt{{w}^{4} + 1024}}{4 {w}^{2}}$

$= \frac{\frac{4 {w}^{4}}{\sqrt{{w}^{4} + 1024}} - \frac{2 \left({w}^{4} + 1024\right)}{\sqrt{{w}^{4} + 1024}}}{4 {w}^{2}} = \frac{2 {w}^{4} - 2048}{4 {w}^{2} \sqrt{{w}^{2} + 1024}}$

$= \frac{{w}^{4} - 1024}{2 {w}^{2} \sqrt{{w}^{4} + 1024}}$

Set the derivative equal to $0$.

${w}^{4} - 1048 = 0 \implies w = \sqrt[4]{1024} \implies w = 4 \sqrt{2}$

The derivative does not exist when $w = 0$.

To find the extrema, find the function values for the endpoints of the domain, $0$ and $\infty$, and for the critical value(s), $4 \sqrt{2}$.

Since $0$ and $\infty$ cannot be plugged into $f \left(w\right)$, find the limit as $w$ approaches those values.

${\lim}_{w \rightarrow 0} f \left(w\right) = \infty$

$f \left(4 \sqrt{2}\right) = 4$

${\lim}_{w \rightarrow \infty} f \left(w\right) = \infty$

Since $w = 4 \sqrt{2}$ is the only critical value on the interval, and is a relative minimum, it is also a global minimum and is the smallest width value that fits the parameters. The length is $2 \sqrt{2}$, found using the original area formula.

Note that since the Pythagorean formula I created used $\frac{w}{2}$ and $l$, which means that $w = 4 \sqrt{2}$ is the side being bisected, which forms a square within the rectangle, which is a common theme in optimization problems of this nature.