I've been reading through the book "Linear Programming" - A Concise Introduction by Thomas S.Ferguson and I've came across this idea that:
The function $x_1$ + $x_2$ is constant on lines with slope -1, for example the line $x_1$ + $x_2$ = 1, and as we move this line further from the origin up and to the right, the value of $x_1$ + $x_2$ increases.
Objective function:
$max$ $x_1$ + $x_2$
Set of constraints:
$ x_1 + 2x_2 <= 4 $
$ 4x_1 + 2x_2 <= 12 $
$ -x_1 + x_2 <= 1 $
$ x_1 >= 0 $ and $x_2 >= 0 $
The plot looks like:
Question: Why does the objective function increase on a line with negative slope instead of positive slope. I understand that the function increases as it goes further away from point (0,0) in the ($x_1$, $x_2$) - plane, but I m confused of why should the function increase on a line with negative slope when we can clearly see that every value on that specific line, eg: $x_1$ + $x_2$ = 1 minimizes .
$$ x_1 +x_2 = Q_{max} \rightarrow \frac{x_1}{\sqrt 2} +\frac{x_2}{\sqrt 2} =\frac{Q_{max}}{\sqrt 2} $$
$$ {x_1}{\cos 45^{\circ}} +{x_2}{\sin 45^{\circ}} =\frac{Q_{max}}{\sqrt 2}=p $$
In the above you have a set of parallel lines in the pedal length $p$ /inclination to normal form $(p-\alpha)$ form.
In order to convert straight line
$$ ax+ by = Q_{max} $$
to the above form
$$ \frac{ax}{\sqrt{a^2+b^2}}+\frac{by}{\sqrt{a^2+b^2}}=\frac{Q_{max} }{\sqrt{a^2+b^2}}=p $$ $$ \rightarrow \,{x \cos \alpha}+ { b \sin \alpha}=p $$
Keeping inclination $ -45^{\circ} $ (slope of maximizing line) the same, you slide colored parallel lines to be farthest away from the origin, never leaving the envelope, which makes target choice as the red line shown :
So that is the geometrical procedure to find max objective function , no matter whether it increases or decreases.
$ x_1 +x_2 = 1 $ is pointless because you already (a priori) fixed up a quantity $Q_{max}=1 $ that you are seeking to fix/determine at the end of the exercise ..
From the construction,optimum point is $( 8/3,2/3), Q_{max}\approx 3.3 $