Find extrema of $u=2(x+y)z+xy$, given the constraint $xyz=1/2, x,y,z>0$.
My attempt was to rewrite this as Lagrange function $F(x,y,z,k)=2(x+y)z+xy+k(xyz-1/2)$, then taking partial derivatives:
$\frac{\partial F}{\partial x}=2z+y+kyz$
$\frac{\partial F}{\partial y}=2z+x+kxz$
$\frac{\partial F}{\partial z}=2x+2y+kyx$
$\frac{\partial F}{\partial k}=xyz-1/2$
Solving this system gives two stationary points $A(1,1,1/2) (k=-4)$ and $B(-1,-1,1/2)(k=0)$. Since $B$ has negative $x$ and $y$, only $A$ is left to consider. Next, $\frac{\partial^2 F}{\partial x^2}=\frac{\partial^2 F}{\partial y^2}=\frac{\partial^2 F}{\partial z^2}=0$. Mixed derivatives are:
$\frac{\partial^2 F}{\partial x\partial y}=zk+1=-1$
$\frac{\partial^2 F}{\partial x\partial z}=yk+2=-2$
$\frac{\partial^2 F}{\partial z\partial y}=xk+2=-2$
Then, $d^2F=2(-dxdy-2dxdz-2dzdy)=-2dxdy-4dxdz-4dzdy$. Differentiating the condition: $-2dy=dz+dx$. Plugging that into $d^2F=dx^2-dzdx+2dz^2$. I am not sure what to do at this point, since I don't see any transformation that would tell me more about nature of total differntial. Any help would be highly appreciated.
It helps to substitute $z=\frac 1 {2xy}$ to write the function as $\frac 1 x +\frac 1y+xy$. It is easy too see that this function is not bounded above. The partial dervatives vanish at $(1,1)$ and the minimum value is $3$.