# Homework Help: Lagrange with Two Constraints

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1. Feb 26, 2017

### Kaura

1. The problem statement, all variables and given/known data

2. Relevant equations

Partials for main equation equal the respective partials of the constraints with their multipliers

3. The attempt at a solution

Basically I am checking to see if this is correct
I am pretty sure that 25/3 is the minimum but I am not sure how to find the maximum
The max an min at the bottom can be ignored or replaced with minimum
I have a lot to do today to prepare for midterms so any help would be much appreciated

2. Feb 26, 2017

### Ray Vickson

Your final solution looks OK, but I did not check the rest because I generally do not look at solutions given as posted images.

You should think about why your solution method does not give you a maximum.

Last edited: Feb 26, 2017
3. Feb 26, 2017

### Kaura

So is 25/3 the only extrema and a minimum?

4. Feb 26, 2017

### Ray Vickson

You tell me. But more importantly, what is the reason?

5. Feb 26, 2017

### Kaura

Yes? because the function is not bound and is continuous?

Last edited: Feb 26, 2017
6. Feb 26, 2017

### Ray Vickson

Right, and because the feasible region (the set of allowed $(x,y,z)$ values) is unbounded. We can find feasible points $(x,y,z)$ with $x,z \to -\infty,\: y \to +\infty$ (and opposite); and of course, $f \to +\infty$ for such points.