# Find local extrema of $x_1+5x_2$ when $2x_1+3x_2=1$ and $x_1-x_2+x_3=0$.

I'm trying to solve the following problem:

Find all local extrema of $f:M\to\mathbb{R}$ where
$$M :=\left\{x\in\mathbb{R}^3: 2x_1+3x_2=1,\ x_1-x_2+x_3=0\right\}$$ and $$f(x)=x_1+5x_2,\ x\in M$$

For sure it is not a demanding problem but I'm already getting a headache trying to think of a solution to this. I understand that this question might not be of the breath-taking one but I would really appreciate any hint. Cheers!

### What I've tried so far:

Well, nothing to be proud of. I've tried applying the Lagrange Multipliers Theorem but I realised I cannot do it because $M$ is not an open set.

I've tried to play a little bit with the conditions from $M$ and see if I can get anything that will reveal extrema of $f$ but I wasn't lucky enough.

Edit: Another thing that is confusing me is that all the variables here have the the power of $1$ so if I get a derivative of any of the equations here I'll get a number, not a function. It makes everything even more difficult since derivatives help a lot with finding extrema usually.

But there are geometrical considerations: Your "manifold" $M$ is the intersection of two nonparallel planes, whence a line in ${\mathbb R}^3$. On the other hand, the objective function $f$ is a linear function in ${\mathbb R}^3$. Then there are two possibilities: Either (i) $f$ grows linearly along the infinite line $M$, whence there is no local extremum, or (ii) $f$ happens to be constant on $M$. In order to find out which of these two is the case you should in a first step determine the direction of the line $M$.