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Graphical Method of Solving Linear Programming Problems

Minimum value...

Question

Minimum value of objective function $z = 2.5 x + y$ subject to the constraints: $x \geq 0, y \geq 0, 3 x + 5 y \leq 15, 5 x + 2 y \leq 10$ is

Does not exist

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20

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5

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0

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Solution

The correct option is D $0$
Given, $z = 2.5 x + y$ subject to the constraints: $x \geq 0, y \geq 0, 3 x + 5 y \leq 15, 5 x + 2 y \leq 10$
$\Rightarrow \frac{x}{5} + \frac{y}{3} \leq 1, \frac{x}{2} + \frac{y}{5} \leq 1$
We get intersection points as $(\frac{20}{19}, \frac{45}{19})$
Now, values of objective function

Corner points $z = 2.5 x + y$

$(2, 0)$ $5$

$(0, 3)$ $3$

$(\frac{20}{19}, \frac{45}{19})$ $5$

$(0, 0)$ $0$ (Minimum)

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