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(x0, y0) and Q(x1, y1) 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(x0, y0). Then, max[F(x, y)] = Ax0 + By0.
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 Axk + Byk; 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(x0, y0), ∴ y0 = – (A/B) x0 + p
i.e. p = y0 + (A/B) x0
Similarly, q = y1 + (A/B) x1
p < q => [y0 + (A/B)x0] < [y1 + (A/B)x1]
i.e. Ax0 + By0 > Ay1 + By1 [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?)