Express material cost as function of radius

In summary, the conversation discusses the cost of a cylindrical can with a volume of 500 cubic centimeters and the materials used to make it. The cost is expressed as a function of the radius and the minimum value of the cost is found through differentiation. The conversation also brings up the methods used to optimize functions in the class.
  • #1
kbillsy
2
0
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?
 
Mathematics news on Phys.org
  • #2
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).
 
  • #3
Are you using the calculus to optimize functions in this class, or are you using graphical means to approximate the critical values?
 

FAQ: Express material cost as function of radius

What is the formula for expressing material cost as a function of radius?

The formula for expressing material cost as a function of radius is: Material Cost = (π * radius^2) * cost per unit area.

How do you calculate the cost per unit area?

The cost per unit area is calculated by dividing the total cost of the material by its total surface area.

What is the significance of using radius as a variable in the function?

Using radius as a variable in the function allows for a more accurate representation of the material cost, as it takes into account the varying size of the material being used.

Can this function be used for any shape or only for circular shapes?

This function can only be used for circular shapes, as the formula for surface area (π * radius^2) is specific to circles. For other shapes, a different formula would need to be used to calculate the surface area.

How can this function be applied in real-world scenarios?

This function can be applied in various real-world scenarios, such as calculating the cost of materials for circular objects like pipes or plates. It can also be used in engineering and construction projects to estimate the cost of circular structures. Additionally, it can be used in manufacturing to determine the cost of circular components.

Back
Top