CBSE 2025, Set 1

Question

For a Linear Programming Problem (LPP), the given objective function is $Z = x + 2y$. The feasible region PQRS determined by the set of constraints is shown as a shaded region in the graph. 

$P \equiv (\frac{3}{13}, \frac{24}{13})$, $Q \equiv (\frac{3}{2}, \frac{15}{4})$, $R \equiv (\frac{7}{2}, \frac{3}{4})$, $S \equiv (\frac{18}{7}, \frac{2}{7})$

Which of the following statements is correct?
![Diagram LPP Feasible Region](https://www.edvaya.com/images/cbse/mathematics/2025/set1/q14-lpp-graph.webp) A). $Z$ is minimum at $S(\frac{18}{7}, \frac{2}{7})$ B). $Z$ is maximum at $R(\frac{7}{2}, \frac{3}{4})$ C). (Value of $Z$ at P) > (Value of $Z$ at Q) D). (Value of $Z$ at Q) < (Value of $Z$ at R)

Edvaya · Accepted Answer

The correct answer is $Z$ is minimum at $S(\frac{18}{7}, \frac{2}{7})$. To find the maximum and minimum values of the objective function $Z = x + 2y$, we evaluate $Z$ at each of the given corner points of the feasible region. 1. *At $P(\frac{3}{13}, \frac{24}{13})$:* $Z = \frac{3}{13} + 2(\frac{24}{13}) = \frac{3 + 48}{13} = \frac{51}{13} \approx 3.92$ 2. *At $Q(\frac{3}{2}, \frac{15}{4})$:* $Z = \frac{3}{2} + 2(\frac{15}{4}) = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9$ (This is the Maximum) 3. *At $R(\frac{7}{2}, \frac{3}{4})$:* $Z = \frac{7}{2} + 2(\frac{3}{4}) = \frac{7}{2} + \frac{3}{2} = \frac{10}{2} = 5$ 4. *At $S(\frac{18}{7}, \frac{2}{7})$:* $Z = \frac{18}{7} + 2(\frac{2}{7}) = \frac{18 + 4}{7} = \frac{22}{7} \approx 3.14$ (This is the Minimum) Now let's evaluate the options: (A) $Z$ is minimum at $S$: *True*, the minimum value is $22/7$. (B) $Z$ is maximum at $R$: *False*, the maximum is at $Q$. (C) $Z$ at $P > Z$ at $Q$: *False*, $3.92$ is not greater than $9$. (D) $Z$ at $Q < Z$ at $R$: *False*, $9$ is not less than $5$. Therefore, the correct statement is that $Z$ is minimum at $S(\frac{18}{7}, \frac{2}{7})$.

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