# Thread: Finding dimensions of a can

1. ## Finding dimensions of a can

I have never been good at solving word problems as I can never figure out how to transfer that into mathematical terms. I wish I could show some effort as I definitely don't want anyone to "just solve it for me" but I have no clue where to even begin with this. Can someone give me a hint please? Thanks!!

Find the dimensions of the right circular can (of volume 400 cubic centimeters) of minimum cost if the material for the top of the can is twice as expensive as the material for the rest of the can.

2. ## Re: Finding dimensions of a can

So in this problem we want to minimize one thing - cost, subject to a constant constraint - volume. Somehow, we need to relate the two. Luckily, the cost is given in relative terms of the surface area of the can (we're not going to be varying the width of the metal here...) and we know how to express surface area in terms of the height and radius. We also know how to express the volume in terms of the height and radius, so it seems natural to try and relate both of these quantities.

First, write equations to describe the cost, and a general equation to describe the volume of a cylinder. You will have something like this:
Let r = radius of can, h=height of can.
Volume = $V=\pi r^2 h$
Cost = C =2(area of the top)+(rest of the area)= $2(\pi r^2)+(2\pi r h + \pi r^2)$

Now we are given that V=400cm cubed, so let's solve the volume equation in terms of one of the other variables, say h, and substitute this constraint on h into our cost function...

(you asked for a hint, so I will leave you to this part on your own)

Now our cost function C is a function of just one variable, r. We can minimize it by taking the derivative with respect to the radius and checking critical points for a minimum. Once we find the minimum, say $r_0$, we can substitute this back into the equation for volume and solve for the corresponding value for height.