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: 

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

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

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?
The level surfaces of the function $ \ f(x,y,z) = 15x + xy - 4x^{2} - 2y^{2} - z^{2} + 2yz + 7 \ = \ c \ $ prove to be triaxial ellipsoids  (using the determinant method shown here ) with their axes not parallel to the coordinates axes and centered on $ \ (2  ,  1  ,  1 ) \ , $ the critical point found by OGC.  We can show that the maximum value of the function, $ \ f(2,1,1) = 15·2 + 2 - 4·2^2 - 2 - 1 + 2 + 7 \ = \ 22 \ $ is the global maximum of the function by an analysis of these surfaces.
The intersections of a level surface with &quot;constant&quot; planes through the center $ \ (2  ,  1  ,  1 ) \  $ are the ellipses
$$ \mathbf{x  =  2:  } \ \  30 \ + \ 2y \ - \ 16 \ - \ 2y^2 \ - \ z^2 \ + \ 2yz \ + \ 7 \ \ = \  -2y^2 \ + \ 2y \ - \ z^2 \ + \ 2yz \ + \ 21 \ \ = \  \ c \ \ ; $$
$$ \mathbf{y  =  1:  } \ \  15x \ + \ x \ - \ 4x^2 \ - \ 2 \ - \ z^2 \ + \ 2z \ + \ 7 \ \ = \  -4x^2 \ + \ 16x \ - \ z^2 \ + \ 2z \ + \ 5 \ \ = \  \ c \ \ ; $$
$$ \mathbf{z  =  1:  } \ \  15x \ + \ xy \ - \ 4x^2 \ - \ 2y^2 \ - \ 1 \ + \ 2y \ + \ 7 $$ $$ = \  -4x^2 \ + \ 15x \ - \ 2y^2 \ + \ 2y \  + \ xy \ + \ 6 \ \ = \  \ c \ \ . $$
The &quot;cross-section&quot; ellipse in the $ \ y \ = \ 1 \ $ plane does have its axes parallel to the $ \ x-$ and $ \ z-$ axes, so we can simply &quot;complete the squares&quot; to obtain
$$ 4·(x^2 - 4x + 4) \ + \ (z^2  +  2z  +  1) \ \ = \  \ 5 \ - \ c \ + \ 16 \ + \ 1  $$ $$ \Rightarrow \ \ 4·(x-2)^2 \ + \ (z-1)^2 \ \ = \  \ 22 \ - \ c \ . $$
This gives us an ellipse for $ \ c \ &lt; \ 22 \ $ and &quot;shrinks to a point&quot; for $ \ c \ = \ 22 \ . $
The other two cross-sections must be rotated first, applying the standard method (coefficients shown are approximate):
$$ \mathbf{x  =  2:  } \ \ \left[ \tan 2\theta \ = \ \frac{2}{(-2)-(-1)}  =  -2 \right] \   -2.618y^2 \ + \ 1.701y \ - \ 0.382 z^2 \ + \ 1.051z \ + \ 21 \  \ = \ \ c \ \ , $$
$$ \mathbf{z  =  1:  } \ \ \left[ \tan 2\theta \ = \ \frac{1}{(-4)-(-2)}  =  -\frac12 \right]  \  -4.118x^2 \ + \ 14.139x \ - \ 1.882 y^2 \ + \ 5.393y \ + \ 6 \  \ = \ \ c \ \ ; $$
after which we &quot;complete the squares&quot; here (note: the centers of the ellipses are altered by these rotations):
$$ \mathbf{x  =  2:  } \quad 2.618·(y^2 \ - \ 0.650y \ + \ 0.1055) \ + \ 0.382·(z^2 \ - \ 2.751z \ + \ 1.892) $$ $$ = \ \ 21 \ - \ c \ + \ 0.276 \ + \ 0.723 \ \ = \ \ 22 \ - \ c \ \ , $$
$$ \mathbf{z  =  1:  } \quad 4.118·(x^2 \ - \ 3.433x \ + \ 2.947) \ + \ 1.882·(y^2 \ - \ 2.866y \ + \ 2.053) $$ $$ = \ \ 6 \ - \ c \ + \ 12.136 \ + \ 3.863 \ \ = \ \ 22 \ - \ c \ \ . $$
So indeed, all of the cross-sections are ellipses for $ \ c \ &lt; \ 22 \ $ and &quot;collapse&quot; to the point $ \ (2,1,1) \ $ for $ \ c \ = \ 22 \ . $  Hence this is the global maximum for our function.
When there is only one critical point and the critical point is a local maximum, that critical point is global maximum as well.