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?

Let us set up the following variables:

`{(r, "Radius (feet"), (y, "Height of can (feet)"), (A, "Surface Area of can (sq feet)") :}`

We want to vary the radius `r` such that we minimise `A`, ie find a critical point of `(dA)/(dr)` that is a minimum, so we to find a function `A(r)`

Then the volume is fixed:

`pir^2h = 3.456pi`
`:. r^2h = 3.456`
`:. h = 3.456/r^2`

And, The Surface Area is given by:

`A=2pirh + 2pir^2`
`:. A=2pir(3.456/r^2) + 2pir^2`
`:. A=(6.912pi)/r + 2pir^2`

Differentiating wrt `r` gives us;

`:. (dA)/(dr)=(6.912pi)(-1/r^2) + 4pir`
`:. (dA)/(dr)=(-6.912pi)/r^2 + 4pir`

At a critical point, `(dA)/(dr)=0`

`:. (-6.912pi)/r^2 + 4pir = 0`
`:. -6.912pi + 4pir^3 = 0`
`:. r^3 = 1.728`
`:. r = 1.2`

With `r=1.2` we have:

`A=(6.912pi)/1.2 + 2pi(1.2)^2`
`:. A=5.76pi + 2.88pi`
`:. A=8.64pi ~~ 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 `(d^2A)/(dr)^2 > 0` when `r=1.2` Instead I will just use the graph `A=(6.912pi)/r + 2pir^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`