# Simple optimization of cylindrical radius for volume

I'm having trouble solving this simple optimization problem, can't work out where I'm going wrong.

A brewery wants to make a cylindrical aluminium beer can which will hold 375ml. (This means the volume of the can is 375 cm3 ). (Assume that any aluminium used for the joins and tab are not included in the calculations.) Set up an appropriate mathematical model which can be used to calculate the radius of the base of the can if the amount of aluminium used in its construction is to be minimised.

So I think: write an equation for the volume (V) and area (A) based on the radius & height:

$$375 = V = \pi h r^2$$ $$A = 2\pi r^2 + \pi h r^2$$

Let's use the constant V to simply to a single variable:

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

To find the optimal value of r which minimizes A, we take the first derivative and solve for 0:

$$\frac{dA}{dr} = 4\pi r$$ $$4\pi r = 0$$

Therefore the optimal value for r to minimize A is:

$$r=0$$

This is clearly wrong, where have I messed up??

• The surface area should be $2\pi r^2 +h\cdot 2\pi r$. The "curved part" has area height * perimeter of the base. – David Mitra Jan 31 '14 at 13:12

Your equation for surface area is not quite right. You have $$A = 2\pi r^2 + \underbrace{\pi h r^2}_{\large = V}$$ In effect, you summed the area of the top and bottom of the can/cylinder with the volume of the can.
We need the area of the base and the top of the can: $2\cdot \pi r^2$, added to the surface area of the "tubular" part of the can: this is the circumference of the base multiplied by the height of the can: $h\cdot 2\pi r$. Summing gives us $$A = 2\pi r^2 + 2h\pi r$$
When solving for $h$ as a function of $r$, we have $$375 = h\pi r^2 \iff h = \frac{375}{\pi r^2}$$
This gives us that $$A = 2\pi r^2 + 2\left(\frac {375}{\pi r^2}\right)$$
Now calculate $\dfrac {dA}{dr}$, and go from there!