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I'm having problems with this assignment:

$$\begin{array}{rl} \min & x^3 + 2xyz - z^2 \\ \text{subject to} & x^2 + y^2 + z^2 \leq 1 \\ \end{array}$$

Disregarding the constraint, find all points $(x,y,z)$ at which the gradient of the criterion function is zero. What is the criterion value at these points? Provide an argument why all these points are neither (unconstrained) local minima nor local maxima, but saddle points.

My approach: I calculated the gradient: $3 x^2+2 y z$; $2 x z$; $2 x y-2 z$ How can I find the values for $x,y$ and $z$ to let the elements be equal to zero?

Any help would be appreciated!

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If this is a problem in your assignment, please add homework tag. –  user21436 Jan 31 '12 at 10:55
    
Added. Any hints for the assignment? –  hplar Jan 31 '12 at 11:06
    
Do you know about the method of Lagrange multipliers? –  Chris Taylor Jan 31 '12 at 11:11
    
yep, but that's the next sub task. –  hplar Jan 31 '12 at 11:13
    
Oops, I didn't read the question properly. Carry on! –  Chris Taylor Jan 31 '12 at 11:24
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2 Answers

From $f_y=2xz=0$ it follows that (a) $x=0$ or (b) $z=0$. In case (a) from $f_z=2(xy-z)=0$ we deduce $z=0$, and $f_x=3x^2+2yz=0$ is then automatically fulfilled, and in case (b) from $f_x=0$ we deduce that $x=0$, and $f_z=0$ is then automatically fulfilled. All in all we conclude that the points $(x,y,z)\in{\mathbb R}^3$ with $\nabla f(x,y,z)=0$ are exactly the points $(0,y,0)$ with $y\in{\mathbb R}$ arbitrary. (These points were already found by superM.) The value of $f$ at such a point is $0$.

Keep $y_0\in{\mathbb R}$ fixed and consider now the straight line $x\ \mapsto\ (x,y_0,0)$ through the point $P_0:=(0,y_0,0)$. The corresponding pullback of $f$ is given by $$\phi(x)=f(x,y_0,0)=x^3$$ which indicates that the function $f$ assumes positive as well as negative values in the immediate neighborhood of $P_0$. Therefore $f$ has neither a local minimum nor a local maximum at $P_0$. Whether you want to call $P_0$ a "saddle point" is up to you; in any case the critical points $(0,y,0)$ of $f$ are so-called degenerate critical points (the determinant of the Hessian is $=0$ in these points), as otherwise they would form a discrete set.

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from the third element in gradient we get z=xy. by substituting this value into the first element we can express y by x and obtain the following values: x = 0, z = 0, y can have any value in its scope.

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