# A cylinder has a volume of 300 cubic inches. The top and bottom parts of the cylinder cost $2 per square inch. And the sides of the cylinder cost$6 per square inch. What are the dimensions of the Cylinder that minimize cost based on these constraints?

Jun 21, 2015

The optimum dimensions are $r \approx 5.2$ inch and $h \approx 3.3$ inch

#### Explanation:

We have to minimize the function $P \left(r , h\right) = 2 \pi \cdot {r}^{2} \cdot 2 + 2 \pi \cdot r \cdot h \cdot 6$
$P \left(r , h\right) = 4 \pi {r}^{2} + 12 \pi \cdot r \cdot h$
$P \left(r , h\right) = 4 \cdot \pi \cdot r \left(r + 3 h\right)$

To get rid of one of the variables we use the fact, that the volume of the cylinder is given.

$V = \pi \cdot {r}^{2} \cdot h$ and $V = 300$, so $300 = \pi \cdot {r}^{2} \cdot h$, so we can calculate, that:

$h = \frac{V}{\pi \cdot {r}^{2}}$

$h = \frac{300}{\pi \cdot {r}^{2}}$

Now we can write $P$ as a function of only one variable:

$P \left(r\right) = 4 \cdot \pi \cdot r \cdot \left(r + 3 \cdot \left(\frac{300}{\pi \cdot {r}^{2}}\right)\right)$

$P \left(r\right) = 4 \cdot \pi \cdot r \cdot \left(r + \frac{900}{\pi \cdot {r}^{2}}\right)$

$P \left(r\right) = 4 \pi \cdot {r}^{2} + \frac{3600}{r}$

Now to find the value of $r$ with tht minimal price $P \left(r\right)$ we have to calculate $P ' \left(r\right)$

$P ' \left(r\right) = 8 \pi \cdot r + \left(\frac{- 3600}{r} ^ 2\right)$
$P ' \left(r\right) = 8 \pi \cdot r - \frac{3600}{r} ^ 2$

Now we have to find $r$ for which $P ' \left(r\right) = 0$

$8 \pi r - \frac{3600}{r} ^ 2 = 0$

$\frac{8 \pi {r}^{3} - 3600}{r} ^ 2 = 0$

$8 \pi {r}^{3} - 3600 = 0$

${r}^{3} = \frac{3600}{8 \pi}$

$r \approx 5.2$ inch

Now we have to calculate $h$ using the formula:

$h = \frac{300}{\pi {r}^{2}}$

$h \approx 3.3$ inch.

So finally we get the dimensions with the minimal price:
$r \approx 5.2$ inch and $h \approx 3.3$ inch.