The LPP: Max. \(z=x_{1}+\frac{5}{2}x_{2}\) subject to \(5x_{1}+3x_{2}\leq 15, -x_{1}+x_{2}\leq 1, 2x_{1}+5x_{2}\leq 10\) and \(x_{1},x_{2}\geq 0\) has
Answer & explanation
Correct answer: option 2
The correct answer is Option 2: Infinitely many optimal solutions
To solve this LPP, we first look at the feasible region created by the constraints:
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$5x_1 + 3x_2 \leq 15$
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$-x_1 + x_2 \leq 1$
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$2x_1 + 5x_2 \leq 10$
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$x_1, x_2 \geq 0$
When we plot these, the feasible region is a small, bounded polygon in the first quadrant. To maximize the objective function $z = x_1 + 2.5x_2$, we test the "corner points" where these lines intersect:
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Intersection of constraints (2) and (3): This gives us the point $(\frac{5}{7}, \frac{12}{7})$. Plugging this into our objective function: $z = (\frac{5}{7}) + 2.5(\frac{12}{7}) = \frac{5 + 30}{7} = \frac{35}{7} = \mathbf{5}$
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Intersection of constraints (1) and (3): This gives us the point $(\frac{45}{19}, \frac{20}{19})$. Plugging this into our objective function: $z = (\frac{45}{19}) + 2.5(\frac{20}{19}) = \frac{45 + 50}{19} = \frac{95}{19} = \mathbf{5}$
The maximum value of 5 is achieved at both of these corner points. Since the maximum value of the objective function is attained at two distinct corner points, every point on the line segment joining these points also gives the same maximum value. Hence, the LPP has infinitely many optimal solutions.