# A metal cylindrical can is to have a volume of 3.456pi cubic feet. How do you find the radius and height of the can which uses the smallest amount of metal?

Nov 29, 2016

The smallest area occurs when we have a radius of $1.2$ feet, leading to an area of $8.64 \pi \approx 27.143$ square feet

#### Explanation:

Let us set up the following variables:

$\left\{\left(r , \text{Radius (feet"), (y, "Height of can (feet)"), (A, "Surface Area of can (sq feet)}\right)\right.$

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

Then the volume is fixed:

$\pi {r}^{2} h = 3.456 \pi$
$\therefore {r}^{2} h = 3.456$
$\therefore h = \frac{3.456}{r} ^ 2$

And, The Surface Area is given by:

$A = 2 \pi r h + 2 \pi {r}^{2}$
$\therefore A = 2 \pi r \left(\frac{3.456}{r} ^ 2\right) + 2 \pi {r}^{2}$
$\therefore A = \frac{6.912 \pi}{r} + 2 \pi {r}^{2}$

Differentiating wrt $r$ gives us;

$\therefore \frac{\mathrm{dA}}{\mathrm{dr}} = \left(6.912 \pi\right) \left(- \frac{1}{r} ^ 2\right) + 4 \pi r$
$\therefore \frac{\mathrm{dA}}{\mathrm{dr}} = \frac{- 6.912 \pi}{r} ^ 2 + 4 \pi r$

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

$\therefore \frac{- 6.912 \pi}{r} ^ 2 + 4 \pi r = 0$
$\therefore - 6.912 \pi + 4 \pi {r}^{3} = 0$
$\therefore {r}^{3} = 1.728$
$\therefore r = 1.2$

With $r = 1.2$ we have:

$A = \frac{6.912 \pi}{1.2} + 2 \pi {\left(1.2\right)}^{2}$
$\therefore A = 5.76 \pi + 2.88 \pi$
$\therefore A = 8.64 \pi \approx 27.143$

We should check that this value leads to a minimum (rather than a maximum) area. As the size of the can is finite this should really be intuitive. We could calculate the second derivative and verify that $\frac{{d}^{2} A}{\mathrm{dr}} ^ 2 > 0$ when $r = 1.2$ Instead I will just use the graph $A = \frac{6.912 \pi}{r} + 2 \pi {r}^{2}$

graph{(6.912pi)/x + 2pix^2 [-2, 15, -500, 500]}

Hopefully you can visually confirm that a minimum does indeed occur when $r = 1.2$