Unbounded Feasible Region for Lagrange with Two Constraints

[Kaura]

Homework Statement

Homework Equations

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

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

 

[Ray Vickson]

Homework Statement

Homework Equations

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

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

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.

 

[Kaura]

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

 

[Ray Vickson]

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

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

 

[Kaura]

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

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

 

[Ray Vickson]

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

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.