Question #3498f

Let us set up the following variables:

`{(r, "Radius (m)"), (y, "Height of can (m)"), (C, "Cost of the can ($)") :}`

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

Then the volume is fixed at `20pi` `m^3`:

`pir^2h = 20pi`
`:. r^2h = 20`
`:. h = 20/r^2`

And, the Surface Area are given by:

`"Side" = 2pirh`
`"Top/Bottom"=2pir^2`

So, the total cost is:

`C=2pirh*8 + 2pir^2*10`
`:. C=16pirh + 20pir^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(h)` and find `h` st `(dC)/(dh)=0`).

`:. C=16pir(20/r^2) + 20pir^2`
`:. C=320pi/r + 20pir^2`

Differentiating wrt `r` gives us;

`:. (dC)/(dr)=(320pi)(-1/r^2) + 40pir`
`:. (dC)/(dr)=(-320pi)/r^2 + 40pir`

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

`:. (-320pi)/r^2 + 40pir = 0`
`:. -320+40r^3=0`
`:. r^3=320/40`
`:. r^3=8`
`:. r = 2`

With `r=2` we have:

`C=(320pi)/2 + 20pi*4`
`:. C=160pi + 80pi`
`:. C=240pi ~~ 753.98` (2dp)

And,

`h=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 `(d^2C)/(dr)^2 > 0` when `r=2` Instead I will just use the graph `C=320pi/r + 20pir^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`