Find Critical Points of Hessian Matrix

[azatkgz]

Please,check my solution.
Find critical points of the function $$f(x,y,z)=x^3+y^2+z^2+12xy+2z$$
and determine their types (degenerate or non-degenerate, Morse index for non-
degenerate).

Attempt

$$\frac{df}{dx}=3x^2+12y=0$$

$$\frac{df}{dy}=2y+12x=0$$

$$\frac{df}{dz}=2z+2=0$$

Critical points are at

x=24 y=-144 z=-1

x=0 y=0 z=-1

$$H(f)=\left|\begin{array}{l[cr]}6x&12&0\\12&2&0\\0&0&2\end{array}\right|$$


for x=24

$$det\left|\begin{array}{l[cr]}144&12&0\\12&2&0\\0&0&2\end{array}\right|=288$$ non-degenerate


for x=0


$$det\left|\begin{array}{l[cr]}0&12&0\\12&2&0\\0&0&2\end{array}\right|=-288$$ non-degenerate

 

[Marco_84]

Please,check my solution.
Find critical points of the function $$f(x,y,z)=x^3+y^2+z^2+12xy+2z$$
and determine their types (degenerate or non-degenerate, Morse index for non-
degenerate).

Attempt

$$\frac{df}{dx}=3x^2+12y=0$$

$$\frac{df}{dy}=2y+12x=0$$

$$\frac{df}{dz}=2z+2=0$$

Critical points are at

x=24 y=-144 z=-1

x=0 y=0 z=-1

$$H(f)=\left|\begin{array}{l[cr]}6x&12&0\\12&2&0\\0&0&2\end{array}\right|$$


for x=24

$$det\left|\begin{array}{l[cr]}144&12&0\\12&2&0\\0&0&2\end{array}\right|=288$$ non-degenerate


for x=0


$$det\left|\begin{array}{l[cr]}0&12&0\\12&2&0\\0&0&2\end{array}\right|=-288$$ non-degenerate

keep going... are they minimum, maximum.. saddle points??

 

[azatkgz]

Morse index

for (0,0,-1)

$$det\left|\begin{array}{l[cr]}-\lambda &12&0\\12&2-\lambda &0\\0&0&2-\lambda\end{array}}\right|=0$$

$$(2-\lambda )(-\lambda(2-\lambda)-144)=0$$

$$\lambda_1=2,\lambda_2=1-\sqrt{145},\lambda_3=1+\sqrt{145}$$


for (24,-144,-1)



$$det\left|\begin{array}{l[cr]}144-\lambda &12&0\\12&2-\lambda &0\\0&0&2-\lambda\end{array}}\right|=0$$

$$(2-\lambda )((144-\lambda)(2-\lambda)-144)=0$$

$$\lambda_1=2,\lambda_2=73-\sqrt{5185},\lambda_3=73+\sqrt{5185}$$

Is it right?What we can say about maximum,minimum and saddle points?

 

[Marco_84]

Morse index

for (0,0,-1)

$$det\left|\begin{array}{l[cr]}-\lambda &12&0\\12&2-\lambda &0\\0&0&2-\lambda\end{array}}\right|=0$$

$$(2-\lambda )(-\lambda(2-\lambda)-144)=0$$

$$\lambda_1=2,\lambda_2=1-\sqrt{145},\lambda_3=1+\sqrt{145}$$


for (24,-144,-1)



$$det\left|\begin{array}{l[cr]}144-\lambda &12&0\\12&2-\lambda &0\\0&0&2-\lambda\end{array}}\right|=0$$

$$(2-\lambda )((144-\lambda)(2-\lambda)-144)=0$$

$$\lambda_1=2,\lambda_2=73-\sqrt{5185},\lambda_3=73+\sqrt{5185}$$

Is it right?What we can say about maximum,minimum and saddle points?

i didnt check you're calculus, but find what the sign of eigenvalues mean and you'll get you're answer.

bye