# I Dot product constrained optimization

1. Sep 21, 2016

### thecage411

Problem:

Fix some vector $\vec{a} \in R^n \setminus \vec{0}$ and define $f( \vec{x} ) = \vec{a} \cdot \vec{x}$. Give an expression for the maximum of $f(\vec{x})$ subject to $||\vec{x}||_2 = 1$.

My work:

Seems like a lagrange multiplier problem.

I have $\mathcal{L}(\vec{x},\lambda) = \vec{a} \cdot \vec{x} - \lambda(||x||_2 - 1)$

Then $D_{xi} \mathcal{L}(\vec{x},\lambda) = a_i - 1/2\lambda(\vec{x} \cdot \vec{x})^{-1/2}2x_i = a_i - \lambda x_i/||x|| = 0$. Solving for $x_i$ yields $x_i = a_i||x||/\lambda$
Also $D_{\lambda} \mathcal{L}(\vec{x},\lambda) = -||x|| + 1 = 0,$ so ||x|| = 1.
Plugging that into the above expression I get $x_i=ai/\lambda$.

But this answer doesn't make sense to me. For one, lambda should fall out, right? Also, just thinking about it -- wouldn't we want to set $x_i = 1$ for the max $a_i$ and have all $j\neq i, x_j = 0$, because any deviation from that would be smaller?

2. Sep 21, 2016

### Krylov

I think that is like killing a fly with a cannon ball, as we say. (The problem does not require such a heavy tool for its solution.)

3. Sep 21, 2016

### thecage411

That's fair -- I gave an argument at the end not using the lagrange multiplier. I guess my question is -- why aren't those two approaches matching up?

4. Sep 21, 2016

### mathman

The simple (second) answer is wrong. $x\cdot a$ is maximum, for fixed length x, when x is parallel and in the same direction as a.