Maximizing the volume of a cylinder

[CrosisBH]

Homework Statement

Find the ratio of R (radius) to H(height) that will maximize the volume of a right circular cylinder for a fixed total surface area.

Relevant Equations

$$\frac{\partial f}{\partial q_k} - \frac{d}{dx} \frac{\partial f}{\partial q_k '} + \sum \lambda_k (x) \frac{\partial g_k}{\partial q_k} = 0$$

$$S = 2\pi RH + 2\pi R^2$$
$$V = \pi R^2 H$$

Note this is in our Lagrangian Mechanics section of Classical Mechanics, so I assume he wants us to use Calculus of Variations to solve it.
The surface area is fixed, so that'll be the constraint. Maximizing volume, we need a functional to represent Volume. This was tricky, but my best guess for it is

V = \int dV = \int_{0}^{H} \pi R^2 dH

That way it can be represented by a single integral.
f = R^2
Plugging the stuff into the Euler-Lagrange equation and simplifying I get (I can show my work here if it's needed)

(2\lambda + 1)R + \lambda H = 0

I'm just stuck here. I don't know how to proceed. Calculus of Variations is still very new to me. Thank you!

 

[BvU]

I can show my work here if it's needed

Yes it's needed ... I'd like to see what you use as Lagrangian and what you use as $q_k$ and $g_k$

[edit] You seem to think that ${\mathcal L} = f = R^2$ but that is not correct
Furthermore, the number of constraints is not equal to the number of variables, so $g_k$ and $\lambda_k$ looks strange: The sum is not over $k$ but over the constraints :$\displaystyle{ \sum_i \lambda_i \frac{\partial g_i}{\partial q_k}}$

The $\lambda_i$ are numbers, not functions.

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[edit2] looked at your profile and want to step back somewhat:

$f$ is the function you want to maximize, so here $f(r,h) = V = 2\pi r^2 h$, a function of 2 variables. At an extreme you expect ${\partial f\over \partial h}= {\partial f\over \partial r} = 0$.
Two equations in two unknowns that are not independent.

Instead of eliminating one of the variables, you use the method of Lagrange multipliers and solve
${\partial {\mathcal L} \over \partial h}= {\partial {\mathcal L} \over \partial r} = 0$ with ${\mathcal L} = f - \lambda g$.

$g(r,h) = 0$ is the constraint $S = 2\pi r^2 + 2\pi rh \quad \Rightarrow \quad g(r,h) = 2\pi r^2 + 2\pi rh - S$.

Now we have three equations (the third one is $g=0$) with three unknowns ($r,h,\lambda$) and we can treat $r$ and $h$ as if they were independent.

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