I have a proposed solution (see below) that I am not 100% satisfied. Please comment.
Given that a convex feasible region lying completely inside the first quadrant has n corner points. Among them, we are only interested in $P(x_0, y_0)$ and $Q(x_1, y_1)$ where P is located lower and righter than Q. $F(x, y) = Ax + By$ is the objective function such that $m(L) > 0$ where $L$ is the simplest line of reference that has the equation $Ax + By= 0$.
$L(Q)$ is a line through $Q$ with slope $= m(L)$ and its y-intercept $= q > 0$, say.
$L(P)$ is a line through $P$ with slope $= m(L)$ and its y-intercept $= p > 0$, say.
[For simplicity only, the argument works well if $p < 0$.]
In linear programming practice, if we want to find max $[F(x, y)]$ graphically, we start from $L(Q)$ and move in the direction “South-East” until we find $P(x_0, y_0)$. Then, max $[F(x, y)] = Ax_0 + By_0$.
The effect will be exactly the same as finding which of the lines has the smaller y-intercept. In this case, it is $L(P)$.
This can then be re-phrased as “if $p<q$, then $Ax_0 + By_0 > Ax_k + By_k$; where $k = 1, 2, … , n$”.
$m(L) > 0$ ; where $L: Ax + By = 0$ => we can assume that $A > 0$ and $B < 0$.
[If not, we can use $L’ : –(Ax + By) = 0$ as our line of reference instead.]
$L(P) : y = – (A/B) x + p$
It passes through $P(x_0, y_0)$, ∴ $y_0 = – (A/B) x_0 + p$
i.e. $p = y_0 + (A/B) x_0$
Similarly, $q = y_1 + (A/B) x_1$
$p < q => [y_0 + (A/B)x_0] < [y_1 + (A/B)x_1]$
i.e. $Ax_0 + By_0 > Ay_1 + By_1$ [since $B < 0$]
I am not comfortable with the argument in the preliminaries.
Take a simple example, let $F(x, y) = x – y$.
Then, we draw the line of reference as $x – y = 0$.
If we draw $– x + y = 0$ instead, what are we calculating?
(maximizing $x – y$ / maximizing $y – x$ / minimizing $x – y$ / minimizing $y - x$?)