MHB Express material cost as function of radius

AI Thread Summary
The discussion focuses on expressing the total cost of a cylindrical can as a function of its radius, given specific material costs for the top, bottom, and sides. The cost function is derived from the volume constraint of 500 cubic centimeters, leading to the equation C = 22πr² + 6000/r, expressed in cents. To find the minimum cost, the derivative of the cost function with respect to radius is set to zero. The conversation also touches on whether calculus or graphical methods are being used to find critical values in the algebra class. Understanding these calculations is essential for solving the problem effectively.
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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?
 
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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).
 
Are you using the calculus to optimize functions in this class, or are you using graphical means to approximate the critical values?
 
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