If at a feasible point (x,y) we get p*=d* that means the point is optimal.
Looking at the example:
min (-x-y)
subject to:
xy<=4
0<=x<=4
0<=y<=8
I define X={(x,y)| 0<=x<=4, 0<=y<=8} and looking for the minimum of the Lagrangian (with one lamda for xy-4<=0) and maximum of the dual function (lamda = 0.5) I get x=y=2 which results in p*=-4=d*. But the feasible point x=0.5, y=8 yields p*=-8.5 < -4 !!
What is wrong here?
Consider the introduction of slack variables like $\epsilon$ With this additional variable the probem can be formulated as
$$ L(x,y,\lambda,\epsilon) = -(x+y)+\lambda(x y - 4-\epsilon^2) $$
The stationary pints can be obtained by solving
$$ \left\{ \begin{array}{rcl} \lambda y & = & 1\\ \lambda x & = & 1\\ x y -4 & = & \epsilon\\ \lambda \epsilon & = & 0 \end{array} \right. $$
Solving we get
$x=2,y=2,\lambda = 1/2,\epsilon = 0$ which indicates that the stationary point is at the boundary ($\epsilon = 0$)
but this is not the solution because $\lambda > 0$ indicating that the objective function and the restriction function grow in the same direction hence the problem has an unbounded solution
For a minimum problem with an active restriction we need
$$ -(-1,-1) + \lambda (2,2) = 0 \Rightarrow \lambda < 0 $$