Square LP. Consider the LP $$\text{minimize }c^Tx\\ \text{subject to } Ax \le b$$ with $A$ square and nonsingular. Show that the optimal value is given by $p^* = c^TA^{-1}b$ if $A^{-1}c \le 0$, and $p^{*} = -\infty$ otherwise.
In the solution the problem is transformed to $$\text{minimize }c^TA^{-1}y\\ \text{subject to } y \le b$$
by the change of variables $y=Ax$. Then, it says if $A^{-T}c \le 0$ the optimal solution is $y = b$, with $p^* = c^TA^{-1}b$. Otherwise, the LP is unbounded below.
The question is how the condition $A^{-1}c \le 0$ was derived (I don't understand the underlying logic)? Even if that condition is satisfied why does it imply that the optimal solution is $y = b$, with $p^* = c^TA^{-1}b$?
Let $d=A^{-T}c$, then your objective function is now
$$\min d^Ty=\sum_{i=1}^ n d_i y_i$$
subject to $$y \le b$$
If one of the term $d_i > 0$, we can set $y_i \to -\infty$ while fixing the other $y_j$ and the overall objective function will go to $-\infty$.
If each of the term $d_i \le 0$, then we would want $y_i$ to be as large as possible to reduce the quantity.