The question in finding maximum and minimum using Langrage method.

Question

Let $f(x,y,z)=x+y+z.$

Find the maximum and minimum of $f$ on $\Omega:=\{(x,y,z)\in \mathbb R^n \mid x^2+y^2+z^2=1, x^3+y^3+z^3=0 \}.$

Here is what my textbook says.

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Since $f$ is continuous and $\Omega$ is compact, $f$ has the maximum and minimum on $\Omega$, and the maximum of $f$ on $\Omega$ is the largest value of the extreme values of $f$ on $\Omega$ , the minimum of $f$ on $\Omega$ is the smallest value of the extreme values of $f$ on $\Omega$ . Thus, what we should do is to find the extreme values of $f$ on $\Omega$ by method of Lagrange multiplier, and to compare these values.

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I don't understand the part

"the maximum of $f$ on $\Omega$ is the largest value of the extreme values of $f$ on $\Omega$ , the minimum of $f$ on $\Omega$ is the smallest value of the extreme values of $f$ on $\Omega$ ." ・・・$(\ast)$

I think this statement implicitly suppose that "the absolute maximum is one of the local maximums".

However, is it correct ?

For example, $f(x)=x^2$ defined on compact $[-1,2]$ has maximum $4$ at $x=2,$ but this is not the local maximum.

Could you explain why we can say $(\ast)$ ?

Answer 1

So the problem in using Lagrange is that we need to know that there exists a maximum/minimum in our desired set $\Omega$.

In general this can become really difficult to prove the existence. If the set $\Omega$ is compact then it simplifies the prove a lot.

To be more precise in $\Bbb{R}$ there is a statement telling that a continuous function $f:\Bbb{R}\rightarrow \Bbb{R}$ on a compact interval $[a,b]$ is always bounded and has a maximum resp. minimum in the interval. This statement can be generalized into more dimensions, I hope you have done this in lecture.

So using this you know that since $f$ is continuous on your set $\Omega$ there exists indeed a maximum resp. minimum. But then what does this mean?

$M\in \Omega$ is a maximum of $f$ if $\forall \omega \in \Omega$ $f(\omega)\leq f(M)$. Similarly for the minimum, so $m\in \Omega$ is a minimum of $f$ if $\forall \omega\in \Omega$ $f(m)\leq f(\omega)$.

But then if there are also further extrem values $E\in \Omega$ this still means that $f(m)\leq f(E)\leq f(M)$. Here you need to pay attention that we have $\leq$ and not $<$. So other extrem values could also be a maximum or minimum.

So yes, the "absolute" maximum is then contained in $\{a\in \Omega: f(a) ~\text{is "extreme"}\}$. So if you apply the Lagrange method you only get possible critical values, so points where $f$ is extreme this includes local extremas. Then to determine if this extremas are maximas or minimas you can to use the Hessematrix. Using this you can determine if the critical points are local max./min. or even global max./min.

I hope this clarifies your question. If not please write again.