To solve the following L.P.P.
Minimize $z = 30 x + 20 y$
Subject to the constraint, $x + y \leq 8$ , $x + 2 y \geq 4$ , $6 x + 7 y \geq 12$ , $x \geq 0; y \geq 0$ .

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Solution

$\begin{matrix} M i n i m i z e = z = 30 x + 20 y x + y \leq 8, x + 2 y \geq 4, 6 x + 7 y \geq 12, x + y \geq 0 x + y = 8 p u t x = 0, 1, 2 \Rightarrow y = 8, 7, 6 r e s p e c t i v e l y x + 2 y = 1 p u t x = 0, 2, 4 y = 2, 1, 0 r e s p e c t i v e l y a n d \Rightarrow 6 x + 7 y = 12 \Rightarrow p u t x = 0, y = \frac{12}{7} \Rightarrow p u t x = 1, = \frac{12}{7} \Rightarrow p u t x = 2, y = 0 \end{matrix}$
hence minimum value of number at (0,2)

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