# Multivariable Calculus application

A firm is producing cylindrical containers to contain a given volume. Suppose that the top and bottom are made of a material that is $N$ times as expensive as the material used for the side of the cylinder, which has a cost of $C$ per unit area. Find, in terms of $N$, the ratio of height to base radius of the cylinder that minimises the cost of producing the containers.

I am not too sure how to go about answering this. I was thinking that you might use the volume of a cylinder formula as a function and then get the partial derivative with respect to the radius, and the height. But I would have no idea what to do with these derivatives or even if that is the right thing to do. I am assuming that it is, as it is from a module in multivariable calculus.

A with any problem-solving exercise, we should understand the problem first, which includes understanding what the goal is. Often this is helped by giving names to the quantities we don't yet know, so that we can talk about them:

"cylindrical containers to contain a given volume"
Let this volume be $V$.

"The top and bottom are made of a materal that is $N$ times as expensive as the material for the side, which costs $\$C$per unit area." So the top and bottom must cost$\$NC$ per unit area.

"Find, in terms of N, the ratio of height to base radius of the cylinder..."
Let the height be $h$ and the base radius be $r$. We are supposed to find $h/r$.

"... that minimises the cost of producing the containers."
Let $T$ be the total cost of the container.

Now that we have defined all the variables, we can start writing equations: \begin{align} V &= \pi r^2 h\\ \text{Surface area of side } &= 2\pi r h\\ \text{Area of top and bottom } &= 2 \pi r^2\\ \text{Cost of side } &= NC \cdot 2 \pi r h\\ \text{Cost of top and bottom } &= C \cdot 2 \pi r^2\\ \text{Total cost, } T &= 2NC \pi r h + 2C \pi r^2 \end{align} We have to find the value of $h/r$ that makes $T$ a minimum, taking into account that the volume is fixed at $V$.

Now we understand the problem, we might be able to finish it.

• Great explanation! Thanks a lot! :D Jul 31 '14 at 16:09