# Finding maximum of a function with an ellipse constraint

I'm trying to find the maximum of a function $f = a^T\mu$ where a is a vector when we have a constraint of the form:

$$g(\mu) = n\mu^T S^{-1}\mu - C = 0$$

where C is a fixed constant, $S^{-1}$ is a fixed matrix , $n$ is a number and $\mu$ is a vector of variables. In fact, my goal is to find a $\mu$ that satisfies $g(\mu)$ and at the same time, maximizes f.

I thought of using Lagrange multipliers but I'm stuck and I appreciate if you can help me solve this problem. Below is what I've done so far:

define: $L = f(\mu) - \lambda g(\mu)$. Then we have:

$\frac{\partial L}{\partial \mu} = a^T - 2 \lambda n \mu^TS^{-1} = 0 \rightarrow \mu^T = \frac{a^TS}{2\lambda n}$

$\frac{\partial L}{\partial \lambda} = g(\mu) = 0$

I don't know how to proceed. Like I don't know how to get rid of lambda in the expression $\mu^T = \frac{a^TS}{2\lambda n}$ I derived above.

Thanks very much for your help.

To get the dual function, you want to plug in the optimal value of $\mu^*$:

$$D(\lambda) = \max_\mu f(\mu) - \lambda g(\mu) = \frac{a^T S a}{4 n \lambda} + C\lambda$$

Then you minimize this dual function to find the optimal $\lambda$.

$$\frac{\partial D}{\partial \lambda} = -\frac{a^T S a}{4 n \lambda^2} + C$$

$$\lambda^* = \sqrt{\frac{a^T S a}{4 n C}}$$

Plugging this back in to the formula for $\mu^*$ gives:

$$\mu^* = \sqrt{\frac{C}{n (a^T S a)}}Sa$$