Suppose I have an optimization problem of the form:
$$\min \mathbf{d}^{T}\mathbf{y}$$
subject to
$$\mathbf{M}\mathbf{y} \geq \mathbf{d}, \qquad \mathbf{y} \geq 0$$
If a solution $\mathbf{s}$ which satisfies $\mathbf{M}\mathbf{s} = \mathbf{d}$ and $\mathbf{s} \geq 0$, how can it be shown $\mathbf{s}$ is optimal for the above linear programming problem?
Initial thought was to form the dual:
max p'd
p>=0
p'M<=d
Thus weak duality gives d'y <= p'd
And the constraint of the dual gives p'M<=d
so : p'My<=dy
and thus p'd<=dy
Combining the two equalities shows that p'd must equal dy, and thus with strong duality this means that p'd is the optimal solution.
Where did I go wrong, and how does being symmetric fit into the problem?