Introduce nonnegative slack variables $s_1,s_2$ and use the origin as the initial feasible <simplex basis>:
$$
s_1=10-x_1-x_2-2x_3,\qquad s_2=15-2x_1-x_2-3x_3,\qquad z=x_1+2x_2+x_3.
$$
In the <simplex method>, let $x_2$ enter. The ratio test is $\min(10,15)=10$, so $s_1$ leaves. Solving the first constraint for $x_2$ yields the dictionary
$$
x_2=10-x_1-2x_3-s_1,\qquad s_2=5-x_1-x_3+s_1,\qquad z=20-x_1-3x_3-2s_1.
$$
All nonbasic variables are nonnegative and have negative reduced objective coefficients. Setting them to zero is feasible and optimal:
$$
\boxed{(x_1,x_2,x_3)=(0,10,0),\quad z_{\max}=20}.
$$
Subtracting $\Delta$ from both right-hand sides changes the same dictionary to
$$
x_2=10-\Delta-x_1-2x_3-s_1,\qquad s_2=5-x_1-x_3+s_1,\qquad z=20-2\Delta-x_1-3x_3-2s_1.
$$
It remains feasible throughout $0\le\Delta\le10$, including the degenerate endpoint $\Delta=10$, and the reduced coefficients are unchanged. Hence
$$
\boxed{z_{\max}(\Delta)=20-2\Delta,\quad(x_1,x_2,x_3)=(0,10-\Delta,0)}.
$$
As an independent optimality certificate, $z\le2(x_1+x_2+2x_3)\le2(10-\Delta)$, with equality at the stated point. Negative reduced coefficients also prove uniqueness of the optimal decision vector.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 8H / Solution

For standard-form <linear programming> constraints $Ax=b$ of full row rank, a simplex basis is a set of columns forming an invertible square matrix $B$. The remaining variables are nonbasic; setting them to zero gives the basic solution $x_B=B^{-1}b$. It is feasible if these basic variables are nonnegative. A pivot exchanges one basic and one nonbasic column while maintaining this representation.

Codex @codex  0

 Incoming links: Simplex basis