# Lagrange Multipliers Global vs Local

I am wondering whether the above statement is true.
"A necessary condition for the constrained optimization problem to have a GLOBAL min or max is that..."
Should the word local replace global?

I am confused about the method of Lagrange Multipliers as well. When we use this method, we get a bunch of points A1,A2,A3,...An.
If we then compute f(A1),f(A2),...,f(An) and take the largest and smallest, are these guaranteed to be the GLOBAL max and min, respectively? If not, under what conditions would this be the case?

Also, when using the Lagrange Multipliers method, do we still need to check all the boundary points separately?

Could someone explain? I would really appreciate!

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Dick
Homework Helper
Read the statement carefully. It says 'a necessary condition for'. It doesn't say 'a sufficient condition for'. Look those phrases up if you have to.

So before doing every question about Lagrange Multipliers, we must first justify the existence of global max/min, right?

Also, is it a 'sufficient condition for' global max/min or local max/min? In the proof, they only talk about local max/min and never talk about global max/min...yet in the statement of the theorem, they put global...

Dick
Homework Helper
A global min/max is also a local min/max in the context of lagrange multipliers. That's why they say necessary and NOT sufficient. There is NOTHING in there that says you don't need to check whether a local min/max is a global min/max. You still have to do it. Face it. If I had said 'global' in this context, I'd be regretting it. Not because what they say is wrong, but because certain people are taking it the wrong way.

HallsofIvy
Homework Helper
Notice that Dick said "A global min/max is also a local min/max". The other way is not true! A local max/min is not necessarily a global max/min. But since any global max/min must be among the local max/min, it is sufficient to look through all the local max to find the gobal max.

That has nothing to do with "Lagrange multipliers". Surely you did the same thing back in 1 dimensional calculus.

Oh no, I typed something not I really menat...sorry about that. I meant "necessary" not sufficient.

In the statement, is (*) a necessary condition for global max/min or is it a necessary condition for local max/min?
i.e.
Global max/min => condition (*)
Local max/min => condition (*)
Which one is true? The second one seems true to me, but the theorem says the first one is true which I doubt.

"A global min/max is also a local min/max"
From 1st year calculus, say considering f on the interval [a,b], the global max/min can occur at a and/or b which isn't a local min/max, it's an endpoint.
The Lagrange Multiplier method for sure gives the critical points in (a,b), but are the endpoints a and b included as well, or do we always have to check the endpoints separately?

Thanks for clearing my doubts!

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LOCAL max/min at a point => at that point, grad f = (lambda_1)(grad(g_1)) + ... + (lambda_k)(grad(g_k)) for some constants lambda_i

But is the converse true?

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Dick
Homework Helper
No. No. No. No. No. Forget lagrange multipliers. A zero derivative doesn't imply it's a min/max. It could be a saddle point. You've taken calculus. You know this. A local min/max implies zero derivative. The converse is not true. No. No. No. No. Why are you going on about this? Zero derivative is a NECESSARY condition for a local min/max. It's NOT SUFFICIENT. I think I said this ages ago.

No. No. No. No. No. Forget lagrange multipliers. A zero derivative doesn't imply it's a min/max. It could be a saddle point. You've taken calculus. You know this. A local min/max implies zero derivative. The converse is not true. No. No. No. No. Why are you going on about this? Zero derivative is a NECESSARY condition for a local min/max. It's NOT SUFFICIENT. I think I said this ages ago.
Yes, I understand this one. e.g. y=x^3, y'=3x^2, y'(0)=0, so x=0 is critical, but x=0 does not give a local max or a local min.

But is this idea exactly analogous to Lagrange's method?
Does this condition give precisely the "critical points" subject to the constraints g1=0, ..., gk=0? (so that the converse of post #7 is false)

Or is it the case that LOCAL max/min at a point <=> at that point, grad f = (lambda_1)(grad(g_1)) + ... + (lambda_k)(grad(g_k)) for some constants lambda_i?

Dick
Homework Helper
It's exactly analogous. The Lagrange multiplier method gives you points where the gradient is zero (critical points). That doesn't tell you any more about min/max than f'(x)=0 in calculus. It is not the case that <=> holds.

It's exactly analogous. The Lagrange multiplier method gives you points where the gradient is zero (critical points). That doesn't tell you any more about min/max than f'(x)=0 in calculus. It is not the case that <=> holds.
Um...are you sure that the gradient is zero? Shouldn't it be a linear combination of grad(g_i)?

Also, I don't understand why a global min/max is also a local min/max.
e.g. Consider y=x^2 on x E [0,2]
global max=4
global min=0
But neither of these is a local max or min.

Dick
Homework Helper
Same thing. The lagrange multiplier thing just enforces a continuous constraint on the the system. The linear combination of the grads just tells you what the grad is on the constrained submanifold. It does NOT deal with discontinuous constraints like being in [0,2]. YOU HAVE TO FIGURE THOSE OUT. JUST LIKE f'(x)=0 in calculus. HOW MANY TIMES DO I HAVE TO SAY THIS?

HallsofIvy
Homework Helper
Also, I don't understand why a global min/max is also a local min/max.
e.g. Consider y=x^2 on x E [0,2]
global max=4
global min=0
But neither of these is a local max or min.
Then how do you define "local" max or min? Certainly 4 is larger than any other value of x2 near x= 2 in the given set.

Then how do you define "local" max or min? Certainly 4 is larger than any other value of x2 near x= 2 in the given set.
Assume the functions are differentiable.

Local max/min at a => a is a critical value <=> f '(a)=0

But y=x^2, y'=2x, y'(2)=2(2)=4 => not local max/min

HallsofIvy