Chapter 6, Problem 1RE

Problems 1-7 refer to the partially completed table of the six basic solutions of the $e$ -system

$\begin{array}{l}2x_1+5x_2+s_1=32\\ x_1+2x_2+s_2=14\\ \begin{array}{ccccc}& x_1& x_2& s_1& s_2\\ \left(\text{A}\right)& 0& 0& 32& 14\\ \left(\text{B}\right)& 0& 6.4& 0& 1.2\\ \left(\text{C}\right)& 0& 7& -3& 0\\ \left(\text{D}\right)& 16& 0& 0& -2\\ \left(\text{E}\right)& & 0& & 0\\ \left(\text{F}\right)& & & 0& 0\end{array}\end{array}$

In basic solution $\left(\text{B}\right)$, which variables are basic?

To determine

The variables which are basic from the basic solution $\left(\text{B}\right)$ of the $e\text{-system}$ $\begin{array}{c}2x_1+5x_2+s_1=32\\ x_1+2x_2{\text{ +s}}_2=14\end{array}$ given in the table below.

$\begin{array}{ccccc}& x_1& x_2& s_1& s_2\\ \left(\text{A}\right)& 0& 0& 32& 14\\ \left(\text{B}\right)& 0& 6.4& 0& 1.2\\ \left(\text{C}\right)& 0& 7& -3& 0\\ \left(\text{D}\right)& 16& 0& 0& -2\\ \left(\text{E}\right)& & 0& & 0\\ \left(\text{F}\right)& & & 0& 0\end{array}$

Answer to Problem 1RE

The basic variables are $x_2,s_2$.

Explanation of Solution

Consider the given $e\text{-system}$.

$\begin{array}{c}2x_1+5x_2+s_1=32\\ x_1+2x_2{\text{ +s}}_2=14\end{array}$

The variables that have been assigned the value zero are non-basic variables and others that have not been assigned the value zero are termed as basic variables.

Basic solution $\left(\text{B}\right)$ of the $e\text{-system}$ is,

$\begin{array}{cccc}{x}_1& x_2& s_1& s_2\\ 0& 6.4& 0& 1.2\end{array}$

Thus, from the basic solution of the $e\text{-system}$, $x_2,s_2$ have not been assigned the value zero.

Therefore, the basic variables are $x_2,s_2$.