# Cylinder with largest volume

• May 20th 2010, 02:58 AM
Keep
Cylinder with largest volume
Given cylinders of equal surface area, fnd the one with largest Volume.

Attempt:

$V = \pi r^2h$

$A = 2\pi r^2 + 2 \pi rh$

Now I want to use Lagrange multiplier and use the area as a constraint. However, the pi is scaring me. Pi is a constant and so far I have only learnt doing it with variables. I thought of substituting pi with x for example but that is variable. Or should I just treat it as constant and go on with it like this:

$2rh = 4r$ .......................... I
$r^2 = 2r$ ........................... II
$A = 2\pi r^2 + 2 \pi rh$ ........................... III

From equation I, h = 2r.

Substituting in III, we get $4\pi r^3 + 4\pi r^2$, which is the same as $\pi r^3 + \pi r^2$

Now I am stuck here.
• May 20th 2010, 03:43 AM
HallsofIvy
This is very strange. You are doing Calculus yet you say "However, the pi is scaring me. Pi is a constant and so far I have only learnt doing it with variables." Have none of the problems you have done involve "2" or "3"? Those are also constants. Even if you had a problem with, say , "xyz", there is a "1" implied that is a constant.

You then have " $2rh= 4r$" and " $r^2= 2r$". Where did those come from? You say you are using the "Lagrange multiplier method" but if so where is the multiplier? Yes, 2rh= 4r implies h= 2r but did you notice that $r^2= 2r$ immediately implies r= 0 or r= 2 (which is not a solution)? And your third "equation", $2\pi r^2h+ 2\pi r^2+ 2\pi rh$ is not even an equation!

The idea of the "lagrange Multiplier Method" is that any maximum or minimum of F(x,y,z) subject to the constraint G(x,y,z)= constant occurs where $\nabla F= \lambda\nabla G$ for some constant $\lambda$.

Here, F is $\pi r^2h$, the volume function, and G is $\pi r^2+ 2\pi rh$, the total surface area.

$\nabla F= 2\pi rh\vec{i}+ \pi r^2\vec{j}$, where I have arbitrarily written $\vec{i}$ for a unit vector in the "r direction" and $\vec{j}$ for a unit vector in the "h direction".

$\nabla G= (2\pi r+ 2\pi h)\vec{i}+ 2\pi r\vec{h}$.

$\nabla F= \lambda \nabla G$ is, then,
$2\pi rh\vec{i}+ \pi r^2\vec{j}= (2\pi r+ 2\pi h)\vec{i}+ 2\pi r\vec{h}$

Setting the components equal we have the two equations
$2\pi rh= 2\lambda\pi (r+ h)$ and $\pi r^2= 2\pi\lambda r$.

Those, together with the equation $\pi r^2+ 2\pi rh= C$, give three equations to solve for r, h, and $\lambda$.

Since it is not necessary to find $\lambda$ to solve this problem, it is often simplest to eliminate $\lambda$ by dividing one equation by another:
$\frac{2\pi rh}{\pi r^2}= \frac{2\pi (r+h)}{2\pi r}$
which reduces to
$\frac{2h}{r}= \frac{r+h}{r}$.

Solve that, together with $\pi r^2+ 2\pi rh= C$.
• May 21st 2010, 03:17 PM
Keep
Thanks. Yeah, you are right, it is really strange. I should have seen pi just like any other constant. Anyway, here is what I should have done:

$V =\pi r^2 h$
$A = \pi r^2 + \pi rh$
$
C = \pi r^2 + \pi rh$

Then using Lagrange multiplier (which as you pointed out wasn't even there in my first post!)

$
2\pi rh = 2\pi r\lambda$

$
\pi r^2 = \pi r\lambda$

$
C = \pi r^2 + \pi rh$

dividing equation I by 2 we have:

$
\pi rh = \pi r\lambda$

$\pi r^2 = \pi r\lambda$
$C = \pi r^2 + \pi rh$

now we divide equation II by equation I we get $r^2 = rh$ which means $r = +- \sqrt {rh}$ which can also be got from your equation $\frac{2h}{r}= \frac{r+h}{r}$.

Substituting for $r^2 = rh$ in III we get $C= 2\pi rh$ Now substitung for $r = \sqrt {rh}$ we get:

$2\pi rh = \pi rh + h\pi \sqrt {rh}$ which is the same as
$rh = h\sqrt {rh}$

this means $h = \frac{rh} {\sqrt {rh}}$

Am I on the right track so far?