Pls comment as I don't have an answer to this exam practice question.

Question.

For the function f:R^2->R defined by

f(x,y)=y^2lnx-xlnx,

deduce whether f(x,y) is convex, concave or neither about the point (x,y)=(e^{-1},0).



Answer.

I will apply the second derivative test:

f_{xx}=-\frac{y^2}{x^2}-\frac{1}{x}
f_{yy}=2lnx
f_{yx}=f_{xy}=\frac{2y}{x}

At the point (e^{-1},0)

f_{xx}=-e
f_{yy}=-2
f_{yx}=f_{xy}=0

and the Hessian matrix looks like this (apologies for the presentation...)

|-e 0|
|0 -2|

which is a diagonal matrix ->-e and -2 are eigenvalues, and since both are negative, the matrix is negative definite.

Therefore, the function is concave about the point (e^{-1},0). Is this all to it?