Consider the LPP: Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. The optimal feasible solution occurs at Option 1: (6, 0) only Option 2: (0, 3) only Option 3: Neither (6, 0) nor (0, 3) Option 4: Both (6, 0) and (0, 3)

Answer: Both (6, 0) and (0, 3). Solution: The problem is to minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, and x, y ≥ 0. In Linear Programming, the optimal solution always occurs at corner points of the feasible region. Finding the corner points by determining where the constraints intersect with the axes: For constraint 2x + y = 3: - When x = 0: y = 3 → Point (0, 3) - When y = 0: x = 1.5 → Point (1.5, 0) For constraint x + 2y = 6: - When x = 0: y = 3 → Point (0, 3) - When y = 0: x = 6 → Point (6, 0) Both constraints pass through (0, 3). Checking feasibility of (0, 3): 2(0) + 3 = 3 ≥ 3 ✓ 0 + 2(3) = 6 ≥ 6 ✓ x = 0 ≥ 0, y = 3 ≥ 0 ✓ Point (0, 3) is feasible. Checking feasibility of (6, 0): 2(6) + 0 = 12 ≥ 3 ✓ 6 + 2(0) = 6 ≥ 6 ✓ x = 6 ≥ 0, y = 0 ≥ 0 ✓ Point (6, 0) is feasible. Calculating the objective function at each corner point: At (0, 3): Z = 0 + 2(3) Z = 6 At (6, 0): Z = 6 + 2(0) Z = 6 Both points give the same value Z = 6. This occurs because the objective function line x + 2y = 6 is parallel to constraint x + 2y = 6. Every point on the line segment from (0, 3) to (6, 0) gives Z = 6, resulting in multiple optimal solutions. Therefore, the optimal feasible solution occurs at both (6, 0) and (0, 3) with minimum value Z = 6.

Algebra

Linear Programming

Medium

common

Consider the LPP: Minimize $Z = x + 2y$ subject to $2x + y \geq 3$ , $x + 2y \geq 6$ , $x, y \geq 0$ . The optimal feasible solution occurs at

(6, 0) only

(0, 3) only

Neither (6, 0) nor (0, 3)

Both (6, 0) and (0, 3)

✅ Correct Option: 4

The problem is to minimize $Z = x + 2y$ subject to $2x + y \geq 3$ , $x + 2y \geq 6$ , and $x, y \geq 0$ .

In Linear Programming, the optimal solution always occurs at corner points of the feasible region.

Finding the corner points by determining where the constraints intersect with the axes:

For constraint $2x + y = 3$ :

When $x = 0$ : $y = 3$ → Point $(0, 3)$
When $y = 0$ : $x = 1.5$ → Point $(1.5, 0)$

For constraint $x + 2y = 6$ :

When $x = 0$ : $y = 3$ → Point $(0, 3)$
When $y = 0$ : $x = 6$ → Point $(6, 0)$

Both constraints pass through $(0, 3)$ .

Checking feasibility of $(0, 3)$ :

$2(0) + 3 = 3 \geq 3$ ✓

$0 + 2(3) = 6 \geq 6$ ✓

$x = 0 \geq 0$ , $y = 3 \geq 0$ ✓

Point $(0, 3)$ is feasible.

Checking feasibility of $(6, 0)$ :

$2(6) + 0 = 12 \geq 3$ ✓

$6 + 2(0) = 6 \geq 6$ ✓

$x = 6 \geq 0$ , $y = 0 \geq 0$ ✓

Point $(6, 0)$ is feasible.

Calculating the objective function at each corner point:

At $(0, 3)$ :

$Z = 0 + 2(3)$

$Z = 6$

At $(6, 0)$ :

$Z = 6 + 2(0)$

$Z = 6$

Both points give the same value $Z = 6$ . This occurs because the objective function line $x + 2y = 6$ is parallel to constraint $x + 2y = 6$ .

Every point on the line segment from $(0, 3)$ to $(6, 0)$ gives $Z = 6$ , resulting in multiple optimal solutions.

Therefore, the optimal feasible solution occurs at both $(6, 0)$ and $(0, 3)$ with minimum value $Z = 6$ .

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Consider the LPP: Minimize Z=x+2yZ = x + 2yZ=x+2y subject to 2x+y≥32x + y \geq 32x+y≥3, x+2y≥6x + 2y \geq 6x+2y≥6, x,y≥0x, y \geq 0x,y≥0. The optimal feasible solution occurs at

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CUET Mathematics Past Year Questions (Topic-Wise)

Consider the LPP: Minimize $Z = x + 2y$ subject to $2x + y \geq 3$ , $x + 2y \geq 6$ , $x, y \geq 0$ . The optimal feasible solution occurs at