Chapter 3, Problem 37RP

Explanation of Solution

To solve given LP problem:

• The given Linear Programming (LP) is,

${\text{min z =6x}}_{\text{1}}{\text{+2x}}_{\text{2}}$

$\text{s}\text{.t}\text{.}3{\text{x}}_{\text{1}}{\text{+2x}}_{\text{2}}\ge 12$        (1)

$2{\text{x}}_{\text{1}}{\text{+4x}}_{\text{2}}\ge \text{12}$        (2)

${\text{x}}_{\text{2}}\ge \text{1}$        (3)

${\text{x}}_{\text{1}}{\text{, x}}_{\text{2}}\ge \text{0}$

• Substitute the value of “${\text{x}}_{\text{1}}$” as “0” in Equation(1),

$\begin{array}{c}3\times {\text{0+2x}}_{\text{2}}\text{=12}\\ {\text{2x}}_{\text{2}}\text{=12}\\ {\text{x}}_{\text{2}}=\frac{12}{2}\\ {\text{x}}_{\text{2}}=6\end{array}$

• Hence the point “${\text{(x}}_1,{\text{x}}_2)$” is “(0,6)”.
• Substitute the value of “${\text{x}}_2$” as “1” in Equation(1),

$\begin{array}{c}{\text{3x}}_1\text{+2}\times 1\text{=12}\\ {\text{3x}}_1\text{=10}\\ {\text{x}}_1=\frac{10}{3}\\ {\text{x}}_1=3.3\end{array}$

• Hence the point “${\text{(x}}_1,{\text{x}}_2)$” is “(3.3,1)”