# Local Maximum Point; Global Maximum Point

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.

• For 2. Global maximum means that no where you get bigger. Can you prove $f(x,y)\leq2$ because then you would be done – Stan Tendijck May 15 at 13:45
• 3. Just pick one other with the value $2$, that is enough. – Stan Tendijck May 15 at 13:46

## 1 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.