# Maximum possible volume

• Jun 6th 2011, 08:29 AM
Punch
Maximum possible volume
http://i952.photobucket.com/albums/a...g/P5270092.jpg
A circular hollow cone of height h and semi-vertical angle $\displaystyle 30^o$ stands on a horizontal table. An upright cylinder of radius $\displaystyle r$ cm and height $\displaystyle d$ cm just fits into the cone as shown in the diagram.

First part of the question: Show that $\displaystyle d=h-r\sqrt{3}$
Second part: Prove that the maximum possible volume of the cylinder is $\displaystyle \frac{4{\pi}h^3}{81}$

I completed the first part and am facing problems with the second part
• Jun 7th 2011, 09:32 AM
HallsofIvy
The area of a cylinder of radius r and height d is $\displaystyle \pi r^2 d$
• Jun 7th 2011, 10:10 AM
Croat
I guess you know about derivatives else you wouldn't have such a task.
Anyway, use the first part, and write the volume as a function of r.
Then just differentiate it and set it to 0. One solution is 0, but of course since that's impossible take the other.
Plugging it back will give you the maximum for the volume.
• Jun 8th 2011, 10:36 AM
Croat
Would just like to note that something bugged me to solve this without differential calculus...
Since you already know what the maximum is supposed to be, you just have to prove that V(r) can't be bigger than that, so:
$\displaystyle V(r)=r^2\pi (h-r\sqrt{3})>\frac{4h^3\pi}{81}$
must not have real positive solutions (positive because we're talking about length).
Anyway, writing it like this:
$\displaystyle r^3\sqrt{3}-r^2 h+\frac{4h^3}{81}<0$
Now, note how you got a square a root right next to the third power but no root by the square... so the substitution $\displaystyle r=\frac{t}{\sqrt{3}}$ seems plausible but also note that adding h to all this, will get you rid of it too. So let:
$\displaystyle r=\frac{th}{\sqrt{3}}$
Our inequality becomes:
$\displaystyle (\frac{th}{\sqrt{3}})^3\sqrt{3}-(\frac{th}{\sqrt{3}})^2 h+\frac{4h^3}{81}<0$
$\displaystyle t^3-t^2+\frac{4}{27}<0$
And with a little imagination:
$\displaystyle t^3-\frac{8}{27}-t^2+\frac{4}{9}<0$
$\displaystyle (t-\frac{2}{3})(t^2-\frac{2}{3}t-\frac{4}{9})-(t-\frac{2}{3})(t+\frac{2}{3})<0$
And factoring to the end:
$\displaystyle (t-\frac{2}{3})^2(t+\frac{1}{3})<0$
The square is always positive, so for this to be negative t would have to be negative (t<-1/3) but that's impossible. And that's the proof. Note that if the expression were equal to zero i.e. t=2/3 you would get the r for which the volume is at maximum.

In the end, I'd just like to say, I'll get a life one day, it's just I'm too busy right now with math :P.
• Jun 9th 2011, 07:56 AM
Punch
Quote:

Originally Posted by Croat
I guess you know about derivatives else you wouldn't have such a task.
Anyway, use the first part, and write the volume as a function of r.
Then just differentiate it and set it to 0. One solution is 0, but of course since that's impossible take the other.
Plugging it back will give you the maximum for the volume.

i tried to write the volume as a function of r, but couldn't get rid of h.

$\displaystyle V={\pi}r^2(h-r\sqrt{3})$
• Jun 9th 2011, 09:40 AM
Croat
You are not supposed to get rid of h. Do not treat h as a variable, but just like a normal number.
You see that the final result is in terms of h, so you simply assume that you have to get r in terms of h... and you couldn't do that if h weren't in the equation.
• Jun 9th 2011, 07:26 PM
Punch
Quote:

Originally Posted by Croat
You are not supposed to get rid of h. Do not treat h as a variable, but just like a normal number.
You see that the final result is in terms of h, so you simply assume that you have to get r in terms of h... and you couldn't do that if h weren't in the equation.

so i'm assuming that h is a constant and does not vary. But the question doesn't state if h is changing. How did u derive that it doesnt change?
• Jun 10th 2011, 09:32 AM
Croat
Ok, so I guess I said it in a wrong way... it doesn't matter whether h varies or not. You look at the volume as a function of one variable i.e. V(r) not V(r,h).
So the only variable in the function $\displaystyle V(r)$ is r and everything else must be constant (i.e. everything except r is constant in the function).
The reason why you don't look at the volume as a function in both r and h is because the end result is expressed with h.

When you look for a maximal value, it usually never includes the variable e.g.
The maximum of $\displaystyle f(x)=-x^2+1$ is 1, but were the function $\displaystyle f(x)=-x^2+h$ the maximum would be h.
Same principle applies above. Hope it's a little clearer now.