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.

The point $P = ( \frac{3}{13}, \frac{24}{13} )$, $Q = ( \frac{3}{15}, \frac{15}{4} )$, $R = ( \frac{7}{3}, \frac{3}{2} )$, $S = ( \frac{18}{7}, \frac{7}{7} )$. Which of the following statements is correct?

• A. $Z$ is minimum at $S \left( \frac{18}{7}, \frac{7}{7} \right)$
• B. $Z$ is maximum at $R \left( \frac{7}{3}, \frac{3}{2} \right)$
• C. $(\text{Value of } Z \text{ at } P)>(\text{Value of } Z \text{ at } Q)$
• D. $(\text{Value of } Z \text{ at } Q)<(\text{Value of } Z \text{ at } R)$

Hint:

For Linear Programming Problems, always evaluate the objective function at the corner points of the feasible region to find the maximum or minimum values.

Answer 1

The Correct Option is C

We are given the objective function $Z = x + 2y$ and the coordinates of the points $P$, $Q$, $R$, and $S$. To evaluate the objective function at each point, we substitute the values of $x$ and $y$ into $Z = x + 2y$. - At point $P \left( \frac{3}{13}, \frac{24}{13} \right)$, we get: \[ Z_P = \frac{3}{13} + 2 \times \frac{24}{13} = \frac{3}{13} + \frac{48}{13} = \frac{51}{13}. \] - At point $Q \left( \frac{3}{15}, \frac{15}{4} \right)$, we get: \[ Z_Q = \frac{3}{15} + 2 \times \frac{15}{4} = \frac{3}{15} + \frac{30}{4} = \frac{3}{15} + \frac{30}{4} = \frac{3}{15} + \frac{120}{15} = \frac{123}{15} = 8.2. \] - At point $R \left( \frac{7}{3}, \frac{3}{2} \right)$, we get: \[ Z_R = \frac{7}{3} + 2 \times \frac{3}{2} = \frac{7}{3} + 3 = \frac{7}{3} + \frac{9}{3} = \frac{16}{3} = 5.33. \] - At point $S \left( \frac{18}{7}, \frac{7}{7} \right)$, we get: \[ Z_S = \frac{18}{7} + 2 \times 1 = \frac{18}{7} + 2 = \frac{18}{7} + \frac{14}{7} = \frac{32}{7} \approx 4.57. \] Thus, the correct statement is $(\text{Value of } Z \text{ at } P)>(\text{Value of } Z \text{ at } Q)$.