Showing Hessian Determinant Positive Definite

Question

I want to check the second order condition of this problem:

max $xy$

s.t. $x^2 + y^2 = 2$

First I compute the Lagrangian

$L(x, y) = xy-\lambda(x^2 + y^2 -2)$

First order conditions:

$\partial L/\partial x = y - 2\lambda x = 0$

$\partial L/\partial y = x - 2\lambda y = 0$

Solving for $x, y, \lambda$ yields

$\lambda=0.5$ and $x,y=\pm1,\pm1$

However when I calculate the Hessian,

$\partial^2L/\partial x^2 = -2\lambda $

$\partial^2L/\partial y^2 = -2\lambda $

$\partial^2L/\partial x \partial y = 1$

Which yields $\mathrm{det}(H)=4\lambda^2-1=0$

My question: I would think (1,1) is a local maximum but why is $D^2(L)$ not positive definite?

Answer 1

You got two stationary points $(1,1),(-1,-1)$, $\lambda=0.5$ but you should also get two other stationary points: $(1,-1)$ and $(-1,1)$, $\lambda = -0.5$.

Remember to determine if $D^2(H)$ is positive or negative definite on the tangent space, not on the whole $\Bbb R^2$.