Optimization problem - right circular cylinder inscribed in cone

[rxh140630]

Homework Statement

Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a right circular cone of radius R and altitude H

Relevant Equations

$V_{cone}=\frac{\pi}3R^2H$
$V_{cylinder}={\pi}r^2h$

Please I do not want the answer, I just want understanding as to why my logic is faulty.

Included as an attachment is how I picture the problem.

My logic:

Take the volume of the cone, subtract it by the volume of the cylinder. Take the derivative. from here I can find the point that the cone will have minimum volume, which will give me the point where the cylinder is at it's maximum volume.

I do not understand why this logic is faulty. Anyways, using the variable in my attachment:

$V_{min}=\frac{\pi}{3}R^2H-{\pi}r^2h=\frac{\pi}{3}R^2H-{\pi}r^{2}(H-y)=\frac{\pi}{3}R^{2}H-{\pi}Hr^2+{\pi}yr^2$

take derivative of V_min with respect to H

$\frac{\pi}{3}R^2-{\pi}r^2=0 => r=\sqrt{\frac13}R$

but this is wrong.

Is my logic wrong, or is my math wrong? And if my logic is wrong, why? I don't see why minimizing the volume of the cone is not = maximizing the volume of the cylinder.

Please do not give me the answer.

 

[Frodo]

Either method is fine and making the volume of the "remaining cone minus cylinder" a minimum will work. That logic is fine.

It's when you start calculating you go wrong. You are jumping ahead too fast and making mistakes. You need to take simple steps where you know each step is correct.

Drawing a good diagram is half way to solving any problem. Your diagram is not very good - see my suggestion below.

R and H are given as constants so you should not be differentiating with respect to H. There is even a subtle hint: R and H are in upper case.

Also the volume you defined isn't a minimum - it's a volume.

1. So what are the variables?

2. How many are there?

3. Can you simplify it so there is only one variable?

4. Can you now get an expression for the remaining volume in terms of constants R and H; and your variable?

5. Can you now differentiate that volume with respect to the variable you have chosen?

I would have drawn the cone with base downwards, and labelled the cylinder with h for its height, and labelled r and R as the radii. I think it makes things much simpler to understand and helps enormously with Step 3.

I have drawn two other possible cylinders in dotted lines.

Incidentally, you can see by inspection that the cylinder starts with zero volume (when ?), increases in volume (as you vary ?) and then shrinks back to zero volume again (when ?).

 

[rxh140630]

Seriously great reply, thank you. I will try again and report back.

 

[Frodo]

Another point.

If you maximise the volume of the cylinder you just have the one formula to work with - the volume of the cylinder. And it is very easy to see what the variables in the problem are.

If you minimise the volume left in the cone you have to work with the volume of the cylinder as before ... and work with the volume of the cone and subtract them.

Surely it is easier just to calculate the volume of the cylinder alone? You have to work with the volume of the cylinder in both cases. Why go to the bother of working with the volume of the cone and doing the subtraction if you don't need to? The more extra work you do and the more complex the expression you work with the greater the chance of error. Easy is goodness in problems.

Incidentally, thinking about "why do I just need the cylinder? So where does the cone come into things?" should lead you to an insight into the problem.

Putting the arrowheads on the dimensions helps a lot - note how similar dimensions are drawn starting from the same position: the centre line for r and R; and the base for h and H.

Putting similar dimensions together (r with R; h with H) also helps, especially with Step 3. You can write the expression you need for Step 3 by visual inspection of my diagram. Note how much more difficult it is to do that with your diagram.

Note also radii r and R are much easier to use than the diameters d and D. Had the problem been phrased with d and D, put them in the diagram as they are given, but put in r and R as the things to work with.

I cannot emphasise it enough: Drawing a good diagram is half way to solving any problem. Never skimp it. Put in all the data you are given and put in everything you need to solve.

 

[rxh140630]

I'm still actually working on this problem. So far, I have that, for the cylinder, r = h gives you the maximum volume of the cylinder.

How did I get this? Well, the volume of a cylinder is
$V = {\pi}r^2h$
and the perimeter, using Frodo's diagram is:
$P = 4r + 2h$
$h = \frac12P -2r$

so $V = {\pi}r^2[\frac12P -2r]$
$V = \frac{{\pi}}2Pr^2 -2{\pi}r^3$
$V' = {\pi}Pr - 6{\pi}r^2=0$
$P-6r=0$
$\frac16P=r$
$P=\frac46P + 2h$
$\frac26P = 2h$
$\frac16P = h$
$r=h$

I do not know how useful that information will be though.. Now I'm trying to figure out how I can find r in terms of R and h in terms of H. It's been a challenge to say the least...

Am I on the right path?

 

[haruspex]

I'm still actually working on this problem. So far, I have that, for the cylinder, r = h gives you the maximum volume of the cylinder.

How did I get this? Well, the volume of a cylinder is
$V = {\pi}r^2h$
and the perimeter, using Frodo's diagram is:
$P = 4r + 2h$
$h = \frac12P -2r$

so $V = {\pi}r^2[\frac12P -2r]$
$V = \frac{{\pi}}2Pr^2 -2{\pi}r^3$
$V' = {\pi}Pr - 6{\pi}r^2=0$
$P-6r=0$
$\frac16P=r$
$P=\frac46P + 2h$
$\frac26P = 2h$
$\frac16P = h$
$r=h$

I do not know how useful that information will be though.. Now I'm trying to figure out how I can find r in terms of R and h in terms of H. It's been a challenge to say the least...

Am I on the right path?

You are treating the perimeter as a constant. That would only be true if the angle at the apex of the cone were a right angle.
R and H are fixed. If r changes, what happens to h?

 

[Delta2]

I think the key to this problem is to find the relationship between $h$, the height of the cylinder and $r$ the radius of its base. For a given $h$ there is a maximum possible $r(h)$ such that the cylinder does not overgrown outside of the cone. And for a given $r$ there is a maximum possible $h(r)$ such that the same overgrown condition holds.

So find this relationship $h(r)$ (or its inverse r(h)) and then place it in the volume of the cylinder $V=\pi r^2 h(r)$ and you will have only one variable ($r$ in this case) for which you can find the maximum with the method of derivative with respect to $r$.

P.S $h(r)$ will contain except $r$, the constants $R$ and $H$ of the cone, if you determine $h(r)$ correctly.

 

[Frodo]

1. You know the expression for the volume of the cylinder in terms of r and h - that's dead easy.
The problem is it has two variables, h and r so is tricky to differentiate - it needs partial differentiation with respect to both variables which you have probably not yet done. So make it easier.

2. You need to eliminate one variable by expressing it in terms of the other.
So find an expression for r in terms of h (or h in terms of r). This is where the cone comes in - it defines that relationship. In my diagram look at similar triangles. (Had you done the partial differentiation you would still have needed this relationship to simplify what you got - it's swings and roundabouts).

3. Substitute your expression for r back into the expression for the cylinder volume.

4 You now have an expression for the volume of the cylinder in terms of the constants R and H and one variable, h (or one variable, r if you chose to express h in terms of r). This expression can be differentiated with respect to that one variable.

5 Differentiate the volume with respect to h and put to zero.

You now have the value of h making the cylinder a maximum or a minimum volume.

The insight I mentioned above is that the relationship between r and h is defined by the edge of the cone which, in this simple example, is a straight line. Had it been a curved line like a circle, or a parabola defined by a function like $y = x^2$, or a more complex function like $y = ax^3 +bx^2 + cx + d$ the solution method is exactly the same: use the circle or the function to express r in terms of h.

Another variable you could have used is s, the slope distance along the side of the cone where the cylinder touches. You now need to express the cylinder volume in terms of R, H and s, and differentiate with respect to s.

 

[rxh140630]

Thank you so much everyone, I will continue to tackle this problem and report my findings .

 

[rxh140630]

Sorry guys, just bumping this because I have the answer now. I'm disappointed that I didn't see the pattern of similar triangles. I know I shouldn't blame my elementary through high school education but I feel like the Texas education system failed me here. I need to go back to the basic's a bit.

I'm currently busy with school (pursuing a Comp Sci degree) so I hope to edit this post soon with how I got my answers, but yes, it came down to just solving for one variable and then setting the first derivative of the volume function = to 0.