# Constrained Calculus of Variations: maximize volume given fixed surface area

We wish to find a curve $$y : C^1[-a, a]$$, where $$a$$ is an unknown parameter, such that $$y(\pm a) = 0$$, and the volume of the solid of revolution generated by rotating $$y$$ around the $$x$$-axis has maximal volume, given that the surface area is fixed. In other words, we let

$$S(y) = \int_{-a}^a2\pi y\sqrt{1 + (y')^2}dx,\quad V(y) = \int_{-a}^a\pi y^2 dx$$ Then the goal is to find $$y$$ and $$a$$ so that $$V(y)$$ is maximized, subject to $$S(y) = A$$, $$y(\pm a) = 0$$.

I'm expecting to end up with $$y$$ a semicircle, in which case $$a = \frac12\sqrt{\frac A\pi}$$. Solving this using Lagrange multipliers and Euler-Lagrange equations however has proven tedious.

The augmented problem I end up with is that of finding the stationary functions of the functional $$\tilde{J}(y; a) = \int_{-a}^a\left(-\pi y^2 + 2\pi\lambda y\sqrt{1 + (y')^2}\right) dx$$ for some real number $$\lambda$$. If we let $$f(y, y')$$ be the integrand, then the Euler-Lagrange equation (which has been written in an equivalent form given that $$f$$ doesn't depend directly on $$x$$) is $$-f + y'f_{y'} = \text{const}$$ or, letting $$C$$ be our constant, $$\pi y^2 - 2\pi \lambda y\sqrt{1 + (y')^2} + \frac{2\pi\lambda y(y')^2}{\sqrt{1 + (y')^2}} = C$$ Now, one may verify (by plugging it in) that $$y(x) = \sqrt{a^2 - x^2}$$ is a solution to this equation, where $$\lambda = a/2$$ and $$C = 0$$, but for the life of me I cannot solve the equation in general (without the knowledge that the solution is a semicircle a priori).

If we do some rearranging, we can end up with $$(\pi y^2 - C)\sqrt{1 + (y')^2} - 2\pi\lambda y = 0$$ in which case introducing the parametrization $$y'(x) = \tan\theta(x)$$ and rearranging gives $$y = \lambda\left(\cos\theta\pm\sqrt{\cos^2\theta+c}\right)$$ for some other constant $$c$$. We can then sub back in $$y' = \tan\theta$$ to get a separable equation in $$\theta$$, namely: $$d\theta\left(-\lambda\cos\theta\pm\frac{2\cos^2\theta}{\sqrt{\cos^2\theta+c}}\right) = dt$$

According to wolframalpha, however, this does not have an analytic antiderivative in general.

How can I arrive at the desired result?

## 1 Answer

Alright I've figured it out.

The stationary functions of the augmented functional must satisfy not only the Euler-Lagrange equation, but also the additional natural boundary condition $$H[y(\pm a)] = 0$$, where $$H[y] = -f + y'f_{y'}$$ (this can be determined just by taking the Gateaux derivative of $$\tilde{J}(y; a)$$ and setting it equal to zero for all admissible directions).

In this case, the $$H[y]$$ is also just our Euler-Lagrange equation, and so $$H[y(\pm a)] = 0$$ tells us that $$C$$ equals zero, in which case our equation is reduced simply to $$y = 2\lambda\cos\theta$$ after the parametrization $$y' = \tan\theta$$.