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

Question.

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

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

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



Answer.

I will apply the second derivative test:

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

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

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

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

$\displaystyle |-e 0|$
$\displaystyle |0 -2|$

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 $\displaystyle (e^{-1},0)$. Is this all to it?