maximizing $z = 11x_1 + 4x_2 + x_3 +15x_4$
over
$3x_1 + x_2 + 2x_3 +4x_4 \leq 28$
$8x_1 + 2x_2 - x_3 + 7x_4 \leq 50$
with $x_1,x_2,x_3,x_4 \geq 0$
gives us a max of 106 a $(0,4,0,6)$ so $x_1$ and $x_3$ are free, the other two ($x_2,x_4$) are basic variables.
if you change the first solution from $28$ to say $28 + \delta$, determine the range on this $\delta$ s.t. $x_2,x_4$ remain basic variables in an optimal solution, and what would $z$ be then?
My initial intuition is that since the second constraint is fixed but the first is NOT, and that at $(0,4,0,6)$, we have
$3(0) + 4 + 2(0) + 4(6) = 28$ $\leq$ $28$
$\Rightarrow$ $\delta \in [0,\infty)$
but the zero case is the same so we rule that out, hence
$\delta \in (0,\infty)$
But I don't know where to go from here, (im pure math and this is an elective applied course), I want to say this IS my range for the $\delta$, and if it is, what is the new maximum $z$ value?
Thanks to anyone who can answer, also it is from Intro to Linear Programming and Game Theory Chapter 5 Section 5 by Keough and Thie (problem 3, and the example is along it in next photo)
Hint: consider the dual problem $$ \min(28y_1+50y_2)\\ \begin{cases} 3y_1+8y_2\ge 11,\\ y_1+2y_2\ge 4,\\ 2y_1-y_2\ge 1,\\ 4y_1+7y_2\ge 15,\\ \text{all }y_k\ge0. \end{cases} $$ For what $28+\delta$ the same solution $A:(2,1)$ holds? You can draw the level set $(28+\delta)y_1+50y_2=\text{const}$ and see on the picture.