Solve the following L.P.P. by graphical method:
Maximise: $Z = 6 x + 4 y$
Subject to $x \leq 2, x + y \leq 3, - 2 x + y \leq 1, x \geq 0, y \geq 0$

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Solution

To Maximize: $Z = 6 x + 4 y$
Constraints: $x \leq 2$
$x + y \leq 3$
$- 2 x + y \leq 1$
$x \geq 0, y \geq 0$

Plotting the constraints on the graph, we get the following points.

Points $Z = 6 x + 4 y$
$O (0, 0)$ $0 \leftarrow$ Minimum
$A (0, 1)$ $4$
$B (\frac{2}{3}, \frac{7}{3})$ $\frac{40}{3}$
$C (2, 1)$ $16 \leftarrow$ Maximum
$D (2, 0)$ $12$
Hence, at $C \equiv (2, 1)$ , Maximum value of $Z = 6 x + 4 y = 16$ .

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