# Finding smallest dimension through optimisation

• Feb 9th 2011, 12:34 AM
xEnOn
Finding smallest dimension through optimisation
What is the smallest dimension of a cone to get a volume of $48 \pi$?

From the cone formula, I get that $r^2 h = 144$. I know it is an optimisation problem but I couldn't figure out what is the equation that I need to form to get its derivative equals to zero.

I tried to make r in terms of h by having it as $(\frac{12}{\sqrt{h}})^2 \times h = 144$ but it will all cancel out to just 144=144 before I could differentiate anything. (Thinking)
• Feb 9th 2011, 12:57 AM
emakarov
Quote:

Originally Posted by xEnOn
What is the smallest dimension of a cone to get a volume of $48 \pi$?

If you are saying "dimension" in singular, then you need to know from the problem statement which of r, h, or their combination you need to minimize. For example, the problem may have been supposed to say that r = h; then you would have only one variable. As it is, you can make r as small as you want at the expense of h and vice versa.
• Feb 9th 2011, 01:26 AM
xEnOn
According to the answer, I think it should mean dimensions in pural. That's, what is the smallest r and h can be to fit into 144. If it is just singular, I think the smallest of just h or r could even go until 0.001.
• Feb 9th 2011, 02:21 AM
xEnOn
Now I am thinking maybe I should take the derivative of the entire cone formula instead of just the $r^2 h$? But I still cannot get the right answer. :(
• Feb 9th 2011, 02:28 AM
emakarov
Your problem in underspecified. There is a whole range of r and h that fit $r^2h=144$. Until one says when $(r_1,h_1)$ is defined to be smaller than $(r_2,h_2)$ or some other information is given, the question is unanswerable.
• Feb 9th 2011, 03:07 AM
xEnOn
I see...thanks. maybe the question has typo in it. thanks! :)