Express material cost as function of radius

[kbillsy]

Hey guys!
So I am taking an algebra class at a college but I am still in high school. And I have been really struggling with it. This is one of the problems that has me quite confused:
A can in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters. The top and bottom are made of a material that costs 11 cents per square centimeter, while the sides are made of a material that costs 6 cents per square centimeter. Express the total cost C of the material as a function of the radius r of the cylinder. Find the right side of the equation. Express the cost in dollars. For what value of r is the cost C a minimum?

 

[Deveno]

Re: Expressing functions

Let's figure out the cost of the can based on the radius, first.

Say the radius is $r$ centimeters. We have a top and bottom that costs 11 cents per square centimeter, so the cost of the lid and the base is:

$2\ast 11\ast \pi r^2$ cents.

If the height of the can is $h$, the cost of the sides of the can is:

$6\ast 2\pi rh$ cents.

However, we have a problem, here...we don't know what "$h$" is.

To figure out what $h$ is, we need to use the fact that the volume of the can is 500 cubic centimeters. Now, the volume in terms of the radius and height is:

$V = \pi r^2h$

and using $V = 500$, we can solve for $h$:

$500 = \pi r^2 h$, so:

$h = \dfrac{500}{\pi r^2}$.

Now our cost is:

$C = 22\pi r^2 + 12\pi rh = 2\pi r(11r + 6h)$

$= 2\pi r\left(11r + 6\dfrac{500}{\pi r^2}\right)$

$= 22\pi r^2 + \dfrac{6000}{r}$

This is a function purely in $r$, so to minimize it, we solve $\dfrac{dC}{dr} = 0$.

(Note: this expresses $C$ in cents, so to get $C$ in dollars, we have to divide by 100. It's also possible I have made an arithmetic error somewhere, i am prone to those).

 

[MarkFL]

Are you using the calculus to optimize functions in this class, or are you using graphical means to approximate the critical values?