# Optimization of a Cylinder's Height and Radius

• Manni
In summary, the problem involves finding the optimal height and radius for a cylindrical can that has a volume of r^2h and a surface area of 2rh. To solve this, you will need to use the formulas for volume and surface area, and add the areas of the top and bottom, which are circles. From there, you can solve for one variable and plug it into the volume formula to find the maximum volume. There are two methods to find the maximum or minimum of a function of one variable, which can be applied to this problem.

#### Manni

A cylindrical can with height h and radius r is to be used to store vegetarian chilli. It
is to be made with 6 square centimetres of tin. Find the height h and radius r which
maximizes the volume of the can.
Hint: The volume of a cylinder is r2h and the surface area of the side walls of a cylinder
is 2rh. The can will also have a top and a bottom, of course — even veggie chilli spoils
if not sealed completely.

I don't know how to solve this question. The mechanics aren't important to me, I'm more concerned of the method. Can somebody explain their methodology for such a question? Thank you!

What are the formulas for volume and surface area? The formula for volume is given. Part of the formula for surface area is given- as the problem says, you need to add the areas of the top and bottom. Both formulas have two variables, the height and radius of the can. Use the information about surface area to solve for one variable and put it into the formula for volume so you have volume depending on one variable.

Do you know how to find max or min of a function of one variable?
(For this particular problem there are two very different methods.)

I'm also having trouble with this problem. Given your recommendation, I added the area of the top and bottom to the surface area formula. The top and bottom are just circles so here's what I got:

SA = 2*pi*r*h + 2*pi*r^2

If I solve for h and plug into the volume formula (pi*r^2*h), where do I go from here?