Cylindrical Optimization Problem - Please Help

[AustrianSaurkraut]

The Question:
Kyle is designing a cylindrical garbage can. The open-topped can will have a volume of 24 000π cm3. The metal for the bottom of the can costs three times as much as the metal for the sides. Determine the ratio of the height to the radius for the dimensions of the can that will minimize the cost of the material.

I've been stuck on this question as the wording really confuses me and I cannot really determine what the primary equation would be.

So far I've done the following:
V = πr²h
24000π= πr²h
24000/r² = h

From here, I don't really know where to go as the wording is really complicated and I don't really know what to set as the "main equation". Especially because it asks for a ratio. Any help is greatly appreciated.

 

[HallsofIvy]

This is an "open topped can" so you will need metal for the cylindrical sides and the circular bottom. The bottom is a circle with radius r so has area \(\displaystyle \pi r^2\). If we were to cut the cylindrical side down its length and "open it up" we would have a rectangle with one side of length h and the other the circumference of the circle, \(\displaystyle 2\pi r\). So that area is \(\displaystyle 2\pi rh\). We are NOT told specific costs for the metal but we are told that "The metal for the bottom of the can costs three times as much as the metal for the sides." Call the cost of the sides "C" (dollars per square inch perhaps) then the sides cost \(\displaystyle 2\pi rhC\). The bottom cost 3C (per square inch) so a total of \(\displaystyle 3\pi r^2C\). That is, the metal for the can costs \(\displaystyle 2\pi rhC+ 3\pi r^2C\).


To find r and h that will minimize that, take the derivatives with respect to r and h and set them equal to 0. That will give r and h in terms of C. Take a ratio to eliminate C.

 

[lex]

So far I've done the following:
[MATH]V = πr^2h\\ 24000π= πr^2h\\ 24000/r^2 = h \hspace2ex \text{ (1)}[/MATH]

Good. Now get an expression for the Total cost. Take the cost of the material for the sides to be 1 (unit of money) per square cm. Then the cost of the material for the bottom is 3 (units of money) per square cm.
So T(otal cost)[MATH]= 3 \pi r^2 + 2 \pi rh \hspace2ex \text{ (2)}[/MATH]Now from your work above, use equation (1) [MATH]h=24000/r^2[/MATH] and substitute it into the equation for total cost, equation (2), replacing h, so you now have T only in terms of r.
Differentiate T with respect to r and then set this equal to 0. This finds the value of r which makes the cost minimum (clearly it will be a minimum).
Use your equation (1) again to find the corresponding value of h.
Finally, find your ratio.