# Question fa2ee

Oct 24, 2016

$\text{length" = "width" = (931/5)^(1/3)~~5.71" ft}$

$\text{height" = 9310/(931/5)^(2/3) ~~ 28.55" ft}$

#### Explanation:

Let $s$ be the side length of the bottom square of the box, and $h$ be its height. Then the volume of the box is

${s}^{2} h = 931$

$\implies h = \frac{931}{s} ^ 2$

The areas of the bottom, top, and sides rectangular, and thus can be calculated as the product of their side lengths:

${A}_{\text{bot" = A_"top}} = {s}^{2}$

${A}_{\text{side}} = s h = s \left(\frac{931}{s} ^ 2\right) = \frac{931}{s}$

Now we can calculate the total cost $C$ (in cents) as a function of $s$:

C = 15A_"bot" + 10A_"top" + 2.5(4A_"side")#

$= 15 {s}^{2} + 10 {s}^{2} + 10 \left(\frac{931}{s}\right)$

$= 25 {s}^{2} + \frac{9310}{s}$

Our goal is to minimize $C$. Note that we must have the bounds $0 < s < \sqrt{931}$ for our initial equation to hold. To find the extrema of $C \left(s\right) = 25 {s}^{2} + \frac{9310}{s}$, we will find where its derivative is $0$ on that interval:

$C ' \left(s\right) = 2 \left(25 s\right) + 9310 \left(- \frac{1}{s} ^ 2\right) = 50 s - \frac{9310}{s} ^ 2 = 0$

$\implies 50 {s}^{3} - 9310 = 0$

$\implies {s}^{3} = \frac{931}{5}$

$\implies s = {\left(\frac{931}{5}\right)}^{\frac{1}{3}}$

So, the only possible minimum could be at $s = {\left(\frac{931}{5}\right)}^{\frac{1}{3}}$. There is also the chance that the cost could decrease as we approach an endpoint, but it's easy to see that as $\frac{9310}{s} \to \infty$ as $s \to {0}^{+}$, meaning the cost also increases without bound. We can check that $s = \sqrt{931}$ results in a higher cost directly:

$C \left({\left(\frac{931}{5}\right)}^{\frac{1}{3}}\right) = 25 {\left(\frac{931}{5}\right)}^{\frac{2}{3}} + \frac{9310}{\frac{931}{5}} ^ \left(\frac{1}{3}\right) \approx 1773$

$C \left(\sqrt{931}\right) = 25 \left(931\right) + \frac{9310}{\sqrt{931}} \ge 25 \left(931\right) > 1773$

Thus, the cost is minimized with the dimensions:

$\text{length" = "width" = (931/5)^(1/3)~~5.71" ft}$

$\text{height" = 9310/(931/5)^(2/3) ~~ 28.55" ft}$