For the feasible reason of a LPP as shown, if the equation of OA and BC are y - 2x = 0 and y - 2x = 4 respectively than constraints for LPP are Option 1: y ≥ 2x, y - 2x ≤ 4, x ≤ 6, y ≥ 0 Option 2: y ≤ 2x, y - 2x ≤ 4, x ≤ 6, y ≥ 0 Option 3: y ≥ 2x, y - 2x ≥ 4, x ≤ 6, y ≥ 0 Option 4: y ≤ 2x, y - 2x ≥ 4, x ≤ 6, y ≥ 0

Answer: y ≥ 2x, y - 2x ≤ 4, x ≤ 6, y ≥ 0. Solution: Line OA is y - 2x = 0, so on or below the line: y ≤ 2x. Line BC is y - 2x = 4 and feasible region lies below: y - 2x ≤ 4. Combined with x ≤ 6, y ≥ 0, this gives the constraints in option (a).

Applied Mathematics

Linear Programming

Medium

For the feasible reason of a LPP as shown, if the equation of OA and BC are $y - 2x = 0$ and $y - 2x = 4$ respectively than constraints for LPP are

$y \geq 2x, y - 2x \leq 4, x \leq 6, y \geq 0$

$y \leq 2x, y - 2x \leq 4, x \leq 6, y \geq 0$

$y \geq 2x, y - 2x \geq 4, x \leq 6, y \geq 0$

$y \leq 2x, y - 2x \geq 4, x \leq 6, y \geq 0$

✅ Correct Option: 1

Line OA is $y - 2x = 0$ , so on or below the line: $y \leq 2x$ . Line BC is $y - 2x = 4$ and feasible region lies below: $y - 2x \leq 4$ . Combined with $x \leq 6, y \geq 0$ , this gives the constraints in option (a).

2023: 7 June Shift 1

The Solution set of the inequality three $3x + 4y \leq 12$ is :

2023: 7 June Shift 1

A company produces two types of belts A & B with a profit of Rs 2 and Rs 1.50 respectively. Belt of type A needs twice as much time to make as belt type B . The company can produce at the most 1000 belts of type B per day . Material for 800 belts is available per day . At the most , 400 buckles for belt type A and 700 for belts type B are available . Then the appropriate LPP is :

2023: 7 June Shift 1

Maximum value of Z equals to 40x + 50y subject to constraints $3x + y \leq 9$ , $x + 2y \leq 8$ , $x, y \geq 0$ occurs at

2023: 22 May Shift 3

If the objective function for an L.P.P. is $z = 3x + 4y$ and the corner points for unbounded feasible region are (9, 0), (4, 3), (2, 5) and (0, 8), then the minimum value of $z$ occurs at :

2026: 29 May Shift 2

The bounded feasible region of an LPP is shown in the diagram given below.

The objective function $z=3x+10y$ attains optimum values at :

A. Point (0,0)

B. Point (0,4)

C. Point (6,3)

D. Point (5,6)

Choose the correct answer from the options given below:

2023: 22 May Shift 3

If objective function $Z = 20x + 30y$ of an LPP is subject to the constraints $3x + 4y \geq 12$ , $4x + y \geq 4$ , $x \geq 0, y \geq 0$ , then Z has :

(A) Min at (0, 4)

(B) Max at (0, 4)

(D) Max at (4, 0)

(E) Min at $\left(\frac{4}{13}, \frac{36}{13}\right)$

Choose the correct answer from the options given below :

For the feasible reason of a LPP as shown, if the equation of OA and BC are y−2x=0y - 2x = 0y−2x=0 and y−2x=4y - 2x = 4y−2x=4 respectively than constraints for LPP are

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CUET Mathematics Past Year Questions (Topic-Wise)

For the feasible reason of a LPP as shown, if the equation of OA and BC are $y - 2x = 0$ and $y - 2x = 4$ respectively than constraints for LPP are