Finding Critical Points and Local Extrema of a Multivariable Function

[Black Orpheus]

For f(x,y) = x^2 + y^2 + 3xy I need to find the critical points and prove whether or not they are local minima, maxima or saddle points. I thought the only critical point was (0,0) since Df = (2x + 3y, 2y + 3x) = 0. Doesn't this make (0,0) a local min? The reason I doubt this now is because upon constructing the Hessian matrix and finding the eigenvalues, I got lambda = 5 and -1 (which would show that the point is a saddle point). Can someone please help me with this?

 

[0rthodontist]

Why are you finding the eigenvalues? What you want to do with the Hessian is find the determinant. The point is a local minimum.

 

[Black Orpheus]

For the problem I am also supposed to find the eigenvalues of the Hessian. Anyway, I think the Hessian is 2 3,3 2 (first row, second row) which makes the determinant 4-9 or -5. This means the det(Hf(x,y)) < 0, which makes the critical point a saddle (second derivative test for local extrema). Is it really a local min?

 

[0rthodontist]

No, I am wrong, you are right. It's a saddle point.

 

[benorin]

Notice that f(x,y)=x^2 + y^2 + 3xy =(x+y)^2+xy

so apply the change of variables:

u=x+y, v=x-y

which has as its inverse

x=\frac{1}{2}(u+v),y=\frac{1}{2}(u-v)

to get

f\left( \frac{1}{2}(u+v), \frac{1}{2}(u-v)\right) =u^2+\frac{1}{4}(u+v)(u-v) = \frac{5}{4}u^2-\frac{1}{4}v^2

which clearly has no extrema (neither local nor global) and the only ciritical point is the saddle point u=v=0, this corresponds to x=y=0 which is likewise a saddle point for the Jacobian of the transformation is

J = \left| \frac{\partial (u,v)}{\partial (x,y)}\right| = \left| \begin{array}{cc} \frac{\partial u}{\partial x}&amp;\frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}&amp;\frac{\partial v}{\partial y}\end{array}\right| = \left| \begin{array}{cc} 1&amp;1\\ 1&amp;-1\end{array}\right| = 1\cdot 1 - 1\cdot (-1) = 2

and the transformation is then just a rotation and a contraction of the old variables.