Cylinder with largest volume

**Keep** · May 20th 2010, 01:58 AM

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.

**HallsofIvy** · May 20th 2010, 02:43 AM

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

**Keep** · May 21st 2010, 02:17 PM

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?

Thread: Cylinder with largest volume

LinkBack

Thread Tools

Display

Cylinder with largest volume

Similar Math Help Forum Discussions

Comparison between cylinder volume and extruded circular spline volume equal dia

largest possible volume of box

Calc 3 - Find the volume of the largest rectangular box

volume of largest rectangular box

Optimization largest volume

Search Tags