To graph the system of inequalities: 1. x + 2y ≤ 3 2. 3x + 4y ≥ 12 3. x ≥ 0 4. y ≥ 1 We draw a graph of the inequalities. For x + 2y ≤ 3, the line x + 2y = 3 crosses the x-axis at (3, 0) and the y-axis at (0, 1.5). - For 3x + 4y ≥ 12, the line 3x + 4y = 12 cuts the x-axis at the point (4, 0) and y-aRead more
To graph the system of inequalities:
1. x + 2y ≤ 3
2. 3x + 4y ≥ 12
3. x ≥ 0
4. y ≥ 1
We draw a graph of the inequalities.
For x + 2y ≤ 3, the line x + 2y = 3 crosses the x-axis at (3, 0) and the y-axis at (0, 1.5).
– For 3x + 4y ≥ 12, the line 3x + 4y = 12 cuts the x-axis at the point (4, 0) and y-axis at (0, 3).
-x ≥ 0 and y ≥ 1 limits the feasible region to the first quadrant above the line y = 1.
When we plot these constraints, we see that the feasible region is empty because the two lines do not intersect within the given constraints.
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function. 1. For the point (0, 3): Z = 4(0) + 5(3) Z = 15 2. For the point (1, 1): Z = 4(1) + 5(1) Z = 9 3. For the point (3, 0): Z = 4(3) + 5(0) Z = 12 The point (1, 1) is where the miniRead more
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function.
1. For the point (0, 3):
Z = 4(0) + 5(3)
Z = 15
2. For the point (1, 1):
Z = 4(1) + 5(1)
Z = 9
3. For the point (3, 0):
Z = 4(3) + 5(0)
Z = 12
The point (1, 1) is where the minimum value of Z = 9 occurs.
To find the minimum value of Z = 11x + 7y, we substitute the corner points of the feasible region into the equation. For point (0, 3), Z = 11(0) + 7(3) = 21. For point (3, 2), Z = 11(3) + 7(2) = 33. For point (0, 5), Z = 11(0) + 7(5) = 35. Thus, the minimum value of Z is 21 at the point (0, 3). ClicRead more
To find the minimum value of Z = 11x + 7y, we substitute the corner points of the feasible region into the equation.
For point (0, 3), Z = 11(0) + 7(3) = 21.
For point (3, 2), Z = 11(3) + 7(2) = 33.
For point (0, 5), Z = 11(0) + 7(5) = 35.
Thus, the minimum value of Z is 21 at the point (0, 3).
"Objective function". In linear programming, the objective function is a function that must be optimized - either maximized or minimized. The variables of the objective function are the decision variables we seek to determine to achieve an optimal solution. Click for more: https://www.tiwariacademy.Read more
“Objective function”. In linear programming, the objective function is a function that must be optimized – either maximized or minimized. The variables of the objective function are the decision variables we seek to determine to achieve an optimal solution.
In order to see which point is not in the half-plane 2x + 3y - 12 ≤ 0, we put in the coordinates for each point in the inequality. 1. Point (1, 2) 2(1) + 3(2) - 12 = 2 + 6 - 12 = -4 ≤ 0. The point (1, 2) is in the half-plane. 2. Point (2, 1) 2(2) + 3(1) - 12 = 4 + 3 - 12 = -5 ≤ 0. Point (2, 1) liesRead more
In order to see which point is not in the half-plane 2x + 3y – 12 ≤ 0, we put in the coordinates for each point in the inequality.
1. Point (1, 2)
2(1) + 3(2) – 12 = 2 + 6 – 12 = -4 ≤ 0. The point (1, 2) is in the half-plane.
2. Point (2, 1)
2(2) + 3(1) – 12 = 4 + 3 – 12 = -5 ≤ 0. Point (2, 1) lies in the half-plane.
3. For point (2, 3):
2(2) + 3(3) – 12 = 4 + 9 – 12 = 1 > 0. Point (2, 3) does NOT lie in the half-plane.
4. For point (-3, 2):
2(-3) + 3(2) – 12 = -6 + 6 – 12 = -12 ≤ 0. Point (-3, 2) lies in the half-plane.
Thus, the point which does not lie in the half-plane is (2, 3).
The number of solutions of the system of in equations x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1 is
To graph the system of inequalities: 1. x + 2y ≤ 3 2. 3x + 4y ≥ 12 3. x ≥ 0 4. y ≥ 1 We draw a graph of the inequalities. For x + 2y ≤ 3, the line x + 2y = 3 crosses the x-axis at (3, 0) and the y-axis at (0, 1.5). - For 3x + 4y ≥ 12, the line 3x + 4y = 12 cuts the x-axis at the point (4, 0) and y-aRead more
To graph the system of inequalities:
1. x + 2y ≤ 3
2. 3x + 4y ≥ 12
3. x ≥ 0
4. y ≥ 1
We draw a graph of the inequalities.
For x + 2y ≤ 3, the line x + 2y = 3 crosses the x-axis at (3, 0) and the y-axis at (0, 1.5).
– For 3x + 4y ≥ 12, the line 3x + 4y = 12 cuts the x-axis at the point (4, 0) and y-axis at (0, 3).
-x ≥ 0 and y ≥ 1 limits the feasible region to the first quadrant above the line y = 1.
When we plot these constraints, we see that the feasible region is empty because the two lines do not intersect within the given constraints.
Thus, the number of solutions is zero.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-12
Corner points of the feasible region determined by the system of linear constraints are (0,3), (1,1) and (3,0). Let Z = 4x + 5y be the objective function. The minimum value of Z occurs at
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function. 1. For the point (0, 3): Z = 4(0) + 5(3) Z = 15 2. For the point (1, 1): Z = 4(1) + 5(1) Z = 9 3. For the point (3, 0): Z = 4(3) + 5(0) Z = 12 The point (1, 1) is where the miniRead more
To determine the minimum value of Z = 4x + 5y, we input the coordinates for the corner points into the objective function.
1. For the point (0, 3):
Z = 4(0) + 5(3)
Z = 15
2. For the point (1, 1):
Z = 4(1) + 5(1)
Z = 9
3. For the point (3, 0):
Z = 4(3) + 5(0)
Z = 12
The point (1, 1) is where the minimum value of Z = 9 occurs.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-12
If the corner points of the feasible region of an LPP are (0, 3), (3, 2) and (0, 5), then the minimum value of Z = 11x 7 y is
To find the minimum value of Z = 11x + 7y, we substitute the corner points of the feasible region into the equation. For point (0, 3), Z = 11(0) + 7(3) = 21. For point (3, 2), Z = 11(3) + 7(2) = 33. For point (0, 5), Z = 11(0) + 7(5) = 35. Thus, the minimum value of Z is 21 at the point (0, 3). ClicRead more
To find the minimum value of Z = 11x + 7y, we substitute the corner points of the feasible region into the equation.
For point (0, 3), Z = 11(0) + 7(3) = 21.
For point (3, 2), Z = 11(3) + 7(2) = 33.
For point (0, 5), Z = 11(0) + 7(5) = 35.
Thus, the minimum value of Z is 21 at the point (0, 3).
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-12
Variable of the objective function of the linear programming problem are
"Objective function". In linear programming, the objective function is a function that must be optimized - either maximized or minimized. The variables of the objective function are the decision variables we seek to determine to achieve an optimal solution. Click for more: https://www.tiwariacademy.Read more
“Objective function”. In linear programming, the objective function is a function that must be optimized – either maximized or minimized. The variables of the objective function are the decision variables we seek to determine to achieve an optimal solution.
Click for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-12
The point which does not lie in the half – plane 2x + 3y -12 ≤ 0 is
In order to see which point is not in the half-plane 2x + 3y - 12 ≤ 0, we put in the coordinates for each point in the inequality. 1. Point (1, 2) 2(1) + 3(2) - 12 = 2 + 6 - 12 = -4 ≤ 0. The point (1, 2) is in the half-plane. 2. Point (2, 1) 2(2) + 3(1) - 12 = 4 + 3 - 12 = -5 ≤ 0. Point (2, 1) liesRead more
In order to see which point is not in the half-plane 2x + 3y – 12 ≤ 0, we put in the coordinates for each point in the inequality.
1. Point (1, 2)
2(1) + 3(2) – 12 = 2 + 6 – 12 = -4 ≤ 0. The point (1, 2) is in the half-plane.
2. Point (2, 1)
2(2) + 3(1) – 12 = 4 + 3 – 12 = -5 ≤ 0. Point (2, 1) lies in the half-plane.
3. For point (2, 3):
2(2) + 3(3) – 12 = 4 + 9 – 12 = 1 > 0. Point (2, 3) does NOT lie in the half-plane.
4. For point (-3, 2):
2(-3) + 3(2) – 12 = -6 + 6 – 12 = -12 ≤ 0. Point (-3, 2) lies in the half-plane.
Thus, the point which does not lie in the half-plane is (2, 3).
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-12