Target Exam

CUET

Subject

Applied Maths. Section B2

Chapter

Linear Programming

Question:

Consider the LPP

Maximize: Z= 5x+3y subject to constraints

$2x+5y ≤15; 5x +2y ≤ 10; x, y ≥0.$ Which of the following is true ?

A. The feasible region lies in the $1^{st}$ quadrant.

B. The maximum value of Z is 9.

C. The feasible region has 4 corner points.

D. It has three optimal solutions.

E. The maximum value of Z lies at $\left(\frac{20}{19}, \frac{45}{19}\right) $ and $(2, 0)$

Choose the correct answer from the options given below :

Options:

A, B, C only

 B, C, D only

A, C only

A, C, E only

Correct Answer:

A, C only

Explanation:

The correct answer is Option (3) → A, C only 

Statement A (True): The non-negativity constraints $x \geq 0, y \geq 0$ mean the feasible region is in the $1^{st}$ quadrant.

Statement B (False): The maximum value of Z is 9.

  • $Z$ at $(2, 0) = 10$, and $Z$ at $(\frac{20}{21}, \frac{55}{21}) \approx 12.62$.

  • Since both values are greater than 9, Statement B is false.

Statement C (True): The region is bounded by the axes and the two lines, having 4 corner points: $(0,0)$, $(0,3)$, $(2,0)$, and the intersection point $(\frac{20}{21}, \frac{55}{21})$.

Statement D (False): It has three optimal solutions.

  • In this case, the maximum value of $12.62$ occurs only at the unique intersection point $(\frac{20}{21}, \frac{55}{21})$.

  • There is no other corner point or line segment that produces this same maximum value, so there is only one optimal solution.

Statement E (False): The maximum value of Z lies at $\left(\frac{20}{19}, \frac{45}{19}\right) $ and $(2, 0)$. The actual intersection point derived from the constraints $2x + 5y = 15$ and $5x + 2y = 10$ is $(\frac{20}{21}, \frac{55}{21})$, which contradicts the coordinates provided in the statement. Furthermore, the value of $Z$ at this intersection point ($\approx 12.62$) is not equal to the value of $Z$ at $(2, 0)$, which is $10$.