Use the two-stage method to solve.
-Find $x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0$ such that
$$\begin{aligned}
& x_{1}+2 x_{2}+x_{3} \leq 40 \
& 2 x_{1}+x_{2}+x_{3} \geq 26
\end{aligned}$$
And $\mathrm{z}=4 \mathrm{x}_{1}+\mathrm{x}_{2}+2 \mathrm{x}_{3}$ is maximized.
A) $x_{1}=40, x_{2}=0, x_{3}=0, z=160$
B) $x_{2}=20, x_{1}=0, x_{3}=0, z=20$
C) $x_{3}=40, x_{1}=0, x_{2}=0, z=80$
D) $x_{1}=30, x_{3}=15, x_{2}=0, z=135$

Question

Quizplus · Accepted Answer

The correct answer is A because the first constraint is binding and the second constraint is not binding.

Use the Two-Stage Method to Solve x1≥0,x2≥0,x3≥0x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0x1​≥0,x2​≥0,x3​≥0

Use the Two-Stage Method to Solve $x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0$