I have to determine the type of critical points of the function:
$$f(x, y, z)=x^2+y^2+z^2+xy$$
I have done the following:
$$\nabla f=(2x+y, 2y+x, 2z) \\ \nabla f=0 \Rightarrow \\ 2x+y=0 \Rightarrow y=-2x \\ 2y+x=0 \Rightarrow 2(-2x)+x=0 \Rightarrow -4x+x=0 \Rightarrow -3x=0 \Rightarrow x=0 \\ 2z=0 \Rightarrow z=0$$
So, the only critical point of $f$ is the point $(0, 0, 0)$.
Is this correct so far??
How can I determine the type of the critical point, if it is a local maxima or a local minima??
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EDIT:
$$f(x, y, z)=x^2+y^2+z^2+xy \\ =\frac{1}{2}x^2+xy+\frac{1}{2}y^2+z^2+\frac{1}{2}x^2+\frac{1}{2}y^2 \\ =\frac{1}{2}(x^2+2xy+y^2)+z^2+\frac{1}{2}(x^2+y^2) \\ =\frac{1}{2}(x+y)^2+z^2+\frac{1}{2}(x^2+y^2) \geq 0$$
Best Answer
That is correct. One way of determinig the critical point is by completing the square. Since the terms are positive you must have a local minimum(in this case a global minimum). Another way is to examine the determinant of the second derivative. If it is +, the you have a minimum. - is a local maximum...