Solve the linear programming problem by graphical method with the following restrictions.
$3 x + 5 y \leq 5; 5 x + 2 y \leq 10; x \geq 0, y \geq 0$ and maximize $Z = 5 x + 3 y$ .

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Solution

From the graph it is very clear that dark shaded portion $A C D E$ is the common (feasible) region for the given constraints.

From linear programming, we know that the maximum is obtained at the vertices of the feasible region.

So, lets find out $Z = 5 x + 3 y$ values at the vertices,
$A (0, 5) : Z = 5 * 0 + 3 * 5 = 15$
$C (0, 1.68) : Z = 5 * 0 + 3 * 1.68 = 5.04$
$D (1.67, 0) : Z = 5 * 1.67 + 3 * 0 = 8.35$
$E (2, 0) : Z = 5 * 2 + 3 * 0 = 10$

$∴ Z = 15$ is the maximum at $(0, 5)$

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