# Is the local max point also a global max?

Consider the following function $$f$$ of three variables, defined on $$\mathbb{R^{3}}$$: $$f(x,y,z) = 15x + xy - 4x^{2} - 2y^{2} - z^{2} + 2yz + 7$$

Is each of the critical points a local maximum? Is each of them a global maximum? Explain.

I've found the critical points after taking the first-order derivatives $$f_{x} = 15 + y - 8x = 0,$$ $$f_{y} = x - 4y + 2z = 0,$$ $$f{z}=-2z + 2y = 0.$$

So there's only one critical point $$\left ( 2,1,1 \right )$$.

To figure out whether the critical point is a local maximum I've computed the partial derivatives and have obtained:

$$H = \begin{bmatrix} -f_{xx} &f_{xy} &f_{xz} \\ f_{yx}&f_{yy} &f_{yz} \\ f_{zx} &f_{zy} &zz \end{bmatrix}$$

$$\Rightarrow H = \begin{bmatrix} -8 &1 &0 \\ 1&-4 &2 \\ 0 &2 &-2 \end{bmatrix}.$$

After using the Method of Leading principal minors, I've figured out that $$H$$ is negative definite so $$\left ( 2,1,1 \right )$$ is a local maximum.

However, I'm a bit confused about whether this local maximum point is also a global maximum point. I know that for a global maximum point, the first-order condition must be satisfied and $$H$$ (the matrix of partial derivatives) must be negative definite for all $$x, y, z \in \mathbb{R^{3}}$$.

So based on this, can I say that this local max point is also a global max because $$H$$ does not depend on the arguments?

• $f_z'$ should be $-2z+2y$, not $2z+2y$. – manooooh Oct 13 '18 at 23:53
• @manooooh Sorry, it's a typo. Have edited. – OGC Oct 13 '18 at 23:56

• This is false. Consider $f(x,y) = -(x^2+y^2(1+x)^3)$. It has a unique critical point at $(0,0)$ which is a local maximum, but restricted to the line $y=x$, it's a polynomial of odd degree which must be unbounded. – KReiser Oct 14 '18 at 0:04