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?
Best Answer
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 "constant" 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 "cross-section" ellipse in the $ \ y \ = \ 1 \ $ plane does have its axes parallel to the $ \ x-$ and $ \ z-$ axes, so we can simply "complete the squares" 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 \ < \ 22 \ $ and "shrinks to a point" 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 "complete the squares" 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 \ < \ 22 \ $ and "collapse" to the point $ \ (2,1,1) \ $ for $ \ c \ = \ 22 \ . $ Hence this is the global maximum for our function.