Solve the following Linear Programming Problems graphically:
Maximise $Z = 5 x + 3 y$
subject to $3 x + 5 y \leq 15, 5 x + 2 y \leq 10, x \geq 0, y \geq 0$ .

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Solution

solve these two equations
$3 x + 5 y <= 15$ -->1
$5 x + 2 y <= 10$ -->1
We will get $(0, 3), (2, 0), (\frac{20}{19}, \frac{45}{19})$
From the graph the corner points are $(0, 3)$ , $(\frac{20}{19}, \frac{45}{19})$ and $(2, 0)$

and the corresponding values of $Z$ are $9$ , $\frac{235}{19} = 12.36$ , $10$
maximum value is $12.36$

Hence the optimal solution is $12.36$

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Z = 5 x + 3 y

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