Local Maximum Point; Global Maximum Point

maxima-minimamultivalued-functionsmultivariable-calculusoptimization

Given is the function:

$f(x,y)=cos(x)+cos(y)$

Which of the following statements is correct?

1. The function has a local maximum point in $P (0, 0)$

This is correct, because the first order condition is fulfilled and the Hessian matrix $Hf$ is negative definite.

$ Hf (x,y) =\begin{pmatrix} -cos(x)&0\\0&-cos(y)\end{pmatrix} $

$ Hf (0,0) =\begin{pmatrix} -1&0\\0&-1\end{pmatrix} $

Eigenvalues are negative, so $Hf$ is positive definite $\Rightarrow$ The function has a local maximum in $P(0,0)$.

2. The function has a global maximum point in $P(0,0)$

How can I check it?
Intuitively, this is correct, because of the rang of the initial function. But how can I prove it?

I think that I should check if the matrix is negative definite everywhere. Is this correct?

3. The function has a maximum point in $P(0,0)$ and this is the only one maximum point with the function value 2.

This should be false, because the function is periodic, i.e. there are infinitely many maximum points with the value 2. How can I prove it?

Any help is appreciated.

Thanks in advance.

Best Answer

Note that $$-1\le \cos(x) \le 1$$ and $$-1\le \cos(y) \le 1$$ for all $(x,y)$

Adding the inequalities result in $$-2\le \cos(x)+\cos(y) \le2$$ for all $(x,y)$

Thus the local maximum at $(0,0)$ is a global maximum as well because of the above inequality.

As you mentioned the function is periodic and this global maximum is attained infinitely many times.

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.

A picture is tremendously helpful. If you have access to maple, try this:

plots[animate](plot,
  [x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);

Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:

x^2+y^2(1-x)^3

[Math] Finding global minumum maximum

There isn't a global maximum or minimum, at least not when considered on $\mathbb{R}^{2}$. Set $y=0$ and you can easily see that your function neither has an upper nor lower bound.

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