For the linear programing problem, Minimize(Z) = 60x + 30y subject to: 2x - y ≥ -5; 3x + y ≥ 3; 2x - 3y ≤ 12; x, y ≥ 0 the optimal value of z is Option 1: 360 Option 2: 150 Option 3: 60 Option 4: No Minimum value of Z

Answer: 60. Solution: The problem requires minimizing Z = 60x + 30y subject to: 2x - y ≥ -5 3x + y ≥ 3 2x - 3y ≤ 12 x, y ≥ 0 In linear programming, the minimum value occurs at a corner point of the feasible region. Finding the intersection of 3x + y = 3 with the y-axis by setting x = 0: 3(0) + y = 3 y = 3 This gives the point (0, 3). Checking the constraints: 2(0) - 3 = -3 ≥ -5 ✓ 3(0) + 3 = 3 ≥ 3 ✓ 2(0) - 3(3) = -9 ≤ 12 ✓ Finding the intersection of 3x + y = 3 with the x-axis by setting y = 0: 3x + 0 = 3 x = 1 This gives the point (1, 0). Checking the constraints: 2(1) - 0 = 2 ≥ -5 ✓ 3(1) + 0 = 3 ≥ 3 ✓ 2(1) - 3(0) = 2 ≤ 12 ✓ Evaluating Z = 60x + 30y at each corner point: At (0, 3): Z = 60(0) + 30(3) Z = 90 At (1, 0): Z = 60(1) + 30(0) Z = 60 Comparing the values, the minimum occurs at (1, 0). Therefore, the optimal value of Z = 60.

Algebra

Linear Programming

Medium

applied

For the linear programing problem,

$Minimize(Z) = 60x + 30y$

subject to: $2x - y \geq -5; 3x + y \geq 3; 2x - 3y \leq 12; x, y \geq 0$

the optimal value of $z$ is

360

150

No Minimum value of Z

✅ Correct Option: 3

The problem requires minimizing $Z = 60x + 30y$ subject to:

$2x - y \geq -5$

$3x + y \geq 3$

$2x - 3y \leq 12$

$x, y \geq 0$

In linear programming, the minimum value occurs at a corner point of the feasible region.

Finding the intersection of $3x + y = 3$ with the $y$ -axis by setting $x = 0$ :

$3(0) + y = 3$

$y = 3$

This gives the point $(0, 3)$ .

Checking the constraints:

$2(0) - 3 = -3 \geq -5$ ✓

$3(0) + 3 = 3 \geq 3$ ✓

$2(0) - 3(3) = -9 \leq 12$ ✓

Finding the intersection of $3x + y = 3$ with the $x$ -axis by setting $y = 0$ :

$3x + 0 = 3$

$x = 1$

This gives the point $(1, 0)$ .

Checking the constraints:

$2(1) - 0 = 2 \geq -5$ ✓

$3(1) + 0 = 3 \geq 3$ ✓

$2(1) - 3(0) = 2 \leq 12$ ✓

Evaluating $Z = 60x + 30y$ at each corner point:

At $(0, 3)$ :

$Z = 60(0) + 30(3)$

$Z = 90$

At $(1, 0)$ :

$Z = 60(1) + 30(0)$

$Z = 60$

Comparing the values, the minimum occurs at $(1, 0)$ .

Therefore, the optimal value of $Z = 60$ .

2025: 14 May Shift 1

Which one of the following set of constraints does the given shaded region represent?

2026: 25 May Shift 1

For the linear programming problem:

Minimize $z = 2x - 5y$ , subject to the constraints:

$x + 2y \ge 10$ ;

$3x + 4y \le 24$ ;

$x,y \ge 0$ , then optimal value of $z$ is

2026: 21 May Shift 1

The maximum value of $z = 3x + 4y$ subject to the constraints

$x + y \geq 2, 2x + 3y \leq 6, x \geq 0, y \geq 0$ is

2026: 30 May Shift 1

For a LPP, the objective function is $z=4x+3y$ and the feasible region determined by a set of linear constraints is shown in the graph as a shaded portion. Which one of the following statement is true ?

2025: 27 May Shift 1

The corner points of the bounded feasible region determined by a set of constraints in an LPP are $P(0, 5)$ , $Q(3, 5)$ , $R(5, 0)$ and $S(4, 1)$ . If the objective function is $z = ax + 2by$ , where, $a, b > 0$ , then the condition on $a$ and $b$ such that the maximum value of $z$ occurs at $Q$ and $S$ is

2025: 15 May Shift 2

If the objective function for a linear programming problem (LPP) is $Z = 4x + 5y$ and the corner points of the bounded feasible region are (9, 0), (4, 3), (2, 5), and (0,8), then the minimum value of Z is:

For the linear programing problem, Minimize(Z)=60x+30yMinimize(Z) = 60x + 30yMinimize(Z)=60x+30y subject to: 2x−y≥−5;3x+y≥3;2x−3y≤12;x,y≥02x - y \geq -5; 3x + y \geq 3; 2x - 3y \leq 12; x, y \geq 02x−y≥−5;3x+y≥3;2x−3y≤12;x,y≥0 the optimal value of zzz is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

For the linear programing problem,

$Minimize(Z) = 60x + 30y$

subject to: $2x - y \geq -5; 3x + y \geq 3; 2x - 3y \leq 12; x, y \geq 0$

the optimal value of $z$ is