# Question #3498f

Dec 2, 2016

The smallest cost occurs when we have a radius of $2$m and a height of $5$m, leading to a cost of $$753.98$. #### Explanation: Let us set up the following variables: $\left\{\begin{matrix}r & \text{Radius (m)" \\ y & "Height of can (m)" \\ C & "Cost of the can ($)}\end{matrix}\right.$

We want to vary the radius $r$ such that we minimise $C$, ie find a critical point of $\frac{\mathrm{dC}}{\mathrm{dr}}$ that is a minimum, so we to find a function $C \left(r\right)$

Then the volume is fixed at $20 \pi$ ${m}^{3}$:

$\pi {r}^{2} h = 20 \pi$
$\therefore {r}^{2} h = 20$
$\therefore h = \frac{20}{r} ^ 2$

And, the Surface Area are given by:

$\text{Side} = 2 \pi r h$
$\text{Top/Bottom} = 2 \pi {r}^{2}$

So, the total cost is:

$C = 2 \pi r h \cdot 8 + 2 \pi {r}^{2} \cdot 10$
$\therefore C = 16 \pi r h + 20 \pi {r}^{2}$

And we can eliminate $h$ so that we have $C$ as a function of $r$ alone (equally we could eliminate $r$ and have $C = f \left(h\right)$ and find $h$ st $\frac{\mathrm{dC}}{\mathrm{dh}} = 0$).

$\therefore C = 16 \pi r \left(\frac{20}{r} ^ 2\right) + 20 \pi {r}^{2}$
$\therefore C = 320 \frac{\pi}{r} + 20 \pi {r}^{2}$

Differentiating wrt $r$ gives us;

$\therefore \frac{\mathrm{dC}}{\mathrm{dr}} = \left(320 \pi\right) \left(- \frac{1}{r} ^ 2\right) + 40 \pi r$
$\therefore \frac{\mathrm{dC}}{\mathrm{dr}} = \frac{- 320 \pi}{r} ^ 2 + 40 \pi r$

At a critical point, $\frac{\mathrm{dC}}{\mathrm{dr}} = 0$

$\therefore \frac{- 320 \pi}{r} ^ 2 + 40 \pi r = 0$
$\therefore - 320 + 40 {r}^{3} = 0$
$\therefore {r}^{3} = \frac{320}{40}$
$\therefore {r}^{3} = 8$
$\therefore r = 2$

With $r = 2$ we have:

$C = \frac{320 \pi}{2} + 20 \pi \cdot 4$
$\therefore C = 160 \pi + 80 \pi$
$\therefore C = 240 \pi \approx 753.98$ (2dp)

And,

$h = \frac{20}{4} = 5$

We should check that this value leads to a minimum (rather than a maximum) cost. As the size of the can is finite this should really be intuitive. We could calculate the second derivative and verify that $\frac{{d}^{2} C}{\mathrm{dr}} ^ 2 > 0$ when $r = 2$ Instead I will just use the graph $C = 320 \frac{\pi}{r} + 20 \pi {r}^{2}$

graph{(320pi)/x + (20pi)*x^2 [-5, 5, -100, 1000]}

Hopefully you can visually confirm that a minimum does indeed occur when $r = 2$