[Math] Finding the minimal cost of a cylindrical can

calculusoptimization

In class, we found the dimensions of a rivht circular cylinder (a "can") that has a volume of 1000 cm3 using the minimum possible material.
This assignment changes that problem slightly by seeking the minimum cost for a right circular cylinder whose volume is 1,000 cm3 where the cost of materials for the bottom, top, and side are different.
Suppose the materials for the bottom of your cylinder cost 6 cents per square centimeter, the materials for the top cost 4 cents per square centimeter, and the materials for the side cost 2 cents for square centimeter.
1. write a function for the total cost of the cylinder in terms of its radius (r) and its height (h).
2. Write an equation expressing the 1,000 cm3 volume in terms of the radius and height. Solve your equation for either r or h and substitute the result into your cost function

I am trying to solve the problem, but I cannot get the equation. Please let me know the answers of these questions.

Best Answer

In class, you probably did something like this:

  • the volume of the can is $V = \pi r^2h$, so $\color{blue}{\pi r^2h=1000}$
  • the material needed is $S(r,h)=2\pi r h + 2\pi r^2$

You then solve $\color{blue}{\pi r^2h=1000}$ for $r$ or $h$ (easier) and substitute into the formula for $S(r,h)$, turning it into a function of one variable. Then use the classic method to find its extreme values.

In the new version of the problem, almost everything stays the same: the volume is still 1000 and the method for finding the minimum is also identical. The only thing that changes is that you do not want to simply minimize the surface area $S$, but a certain cost $C$.

Suppose the materials for the bottom of your cylinder cost 6 cents per square centimeter, the materials for the top cost 4 cents per square centimeter, and the materials for the side cost 2 cents for square centimeter.

The total cost is the sum of:

  • a bottom surface of $\pi r^2$ units at 6 cents per unit, so a cost of $\color{red}{6\pi r^2}$;
  • a top surface of $\pi r^2$ units at 4 cents per unit, so a cost of $\color{red}{4\pi r^2}$;
  • a side surface of $2\pi r h$ units at 2 cents per unit, so a cost of $\color{red}{4\pi r h}$.

The total cost $C(r,h)$ is the sum of all the red contributions. Continue like in the first example.

Can you take it from here? And take better care of your questions here next time ;-).

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