The optimal value of the objective function in LPP is always obtained at the corner points of the feasible region. This is because the objective function is linear, and at one of the corner points due to the property of linear programming, the maximum or minimum occurs. Check this for more: https://Read more
The optimal value of the objective function in LPP is always obtained at the corner points of the feasible region.
This is because the objective function is linear, and at one of the corner points due to the property of linear programming, the maximum or minimum occurs.
To find the condition on p and q, we substitute the coordinates of the corner points 1, 1 and 3, 0 into the objective function Z = px + qy. 1. For point (1, 1): Z = p(1) + q(1) = p + q 2. For point (3, 0) Z = p(3) + q(0) = 3p For the minimum value of Z to occur at (3, 0) and (1, 1), the objective fuRead more
To find the condition on p and q, we substitute the coordinates of the corner points 1, 1 and 3, 0 into the objective function Z = px + qy.
1. For point (1, 1):
Z = p(1) + q(1) = p + q
2. For point (3, 0)
Z = p(3) + q(0) = 3p
For the minimum value of Z to occur at (3, 0) and (1, 1), the objective function value at (1, 1) must be greater than at (3, 0). So we require,
p + q ≥ 3p
q ≥ 2p
To solve the LPP, we first plot the constraints: 1. x - y ≥ -1 ⟹ y ≤ x + 1 2. x ≤ 3 3. x ≥ 0 4. y ≥ 0 Now, we find the corner points of the feasible region: - From x = 0 and y = 0, the point is (0, 0). - For the line x = 3 and y = x + 1, when x = 3, y = 4. So the point is (3, 4). - From the intersecRead more
To solve the LPP, we first plot the constraints:
1. x – y ≥ -1 ⟹ y ≤ x + 1
2. x ≤ 3
3. x ≥ 0
4. y ≥ 0
Now, we find the corner points of the feasible region:
– From x = 0 and y = 0, the point is (0, 0).
– For the line x = 3 and y = x + 1, when x = 3, y = 4. So the point is (3, 4).
– From the intersection of x = 3 and y = x + 1, we get (3, 4).
Now, evaluate Z = 3x + 4y at the corner points:
– At (0, 0), Z = 3(0) + 4(0) = 0
– At (3, 4), Z = 3(3) + 4(4) = 9 + 16 = 25
The maximum value of Z occurs at (3, 4) and is 25.
Given the objective function Z = ax + by, the minimum value occurs at two points (3, 4) and (4, 3). 1. At point (3, 4), the objective function is Z = a(3) + b(4) = 3a + 4b. 2. At point (4, 3), the objective function is Z = a(4) + b(3) = 4a + 3b. Since both points give the minimum value of Z, the objRead more
Given the objective function Z = ax + by, the minimum value occurs at two points (3, 4) and (4, 3).
1. At point (3, 4), the objective function is Z = a(3) + b(4) = 3a + 4b.
2. At point (4, 3), the objective function is Z = a(4) + b(3) = 4a + 3b.
Since both points give the minimum value of Z, the objective function at these points must be equal:
To solve for the maximum value of Z = 3x + 4y subject to the constraints: 1. x ≥ 0 2. y ≥ 0 3. x + y ≤ 1 We plot the constraints first and determine the feasible region. The feasible region is bounded by the points: - (0, 0) where x = 0 and y = 0 - (1, 0) where x + y = 1 and y = 0 - (0, 1) where x +Read more
To solve for the maximum value of Z = 3x + 4y subject to the constraints:
1. x ≥ 0
2. y ≥ 0
3. x + y ≤ 1
We plot the constraints first and determine the feasible region.
The feasible region is bounded by the points:
– (0, 0) where x = 0 and y = 0
– (1, 0) where x + y = 1 and y = 0
– (0, 1) where x + y = 1 and x = 0
Then we calculate Z = 3x + 4y at all the corner points:
– At (0, 0), Z = 3(0) + 4(0) = 0
– At (1, 0), Z = 3(1) + 4(0) = 3
– At (0, 1), Z = 3(0) + 4(1) = 4
The maximum value for Z is 4 and it occurs at (0, 1).
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 optimal value of the objective function is obtained at the points
The optimal value of the objective function in LPP is always obtained at the corner points of the feasible region. This is because the objective function is linear, and at one of the corner points due to the property of linear programming, the maximum or minimum occurs. Check this for more: https://Read more
The optimal value of the objective function in LPP is always obtained at the corner points of the feasible region.
This is because the objective function is linear, and at one of the corner points due to the property of linear programming, the maximum or minimum occurs.
Check this 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 = px + qy, where p, q > 0. Condition on p and q so that minimum value of Z occurs at (3, 0) and (1,1) is
To find the condition on p and q, we substitute the coordinates of the corner points 1, 1 and 3, 0 into the objective function Z = px + qy. 1. For point (1, 1): Z = p(1) + q(1) = p + q 2. For point (3, 0) Z = p(3) + q(0) = 3p For the minimum value of Z to occur at (3, 0) and (1, 1), the objective fuRead more
To find the condition on p and q, we substitute the coordinates of the corner points 1, 1 and 3, 0 into the objective function Z = px + qy.
1. For point (1, 1):
Z = p(1) + q(1) = p + q
2. For point (3, 0)
Z = p(3) + q(0) = 3p
For the minimum value of Z to occur at (3, 0) and (1, 1), the objective function value at (1, 1) must be greater than at (3, 0). So we require,
p + q ≥ 3p
q ≥ 2p
Therefore, the constraint on p and q is q = 2p.
See lessFor the following LPP Maximise Z = 3x + 4y subject to constraints x – y ≥ – 1, x ≤ 3 x ≥ 0, y ≥ 0 The maximum value is
To solve the LPP, we first plot the constraints: 1. x - y ≥ -1 ⟹ y ≤ x + 1 2. x ≤ 3 3. x ≥ 0 4. y ≥ 0 Now, we find the corner points of the feasible region: - From x = 0 and y = 0, the point is (0, 0). - For the line x = 3 and y = x + 1, when x = 3, y = 4. So the point is (3, 4). - From the intersecRead more
To solve the LPP, we first plot the constraints:
1. x – y ≥ -1 ⟹ y ≤ x + 1
2. x ≤ 3
3. x ≥ 0
4. y ≥ 0
Now, we find the corner points of the feasible region:
– From x = 0 and y = 0, the point is (0, 0).
– For the line x = 3 and y = x + 1, when x = 3, y = 4. So the point is (3, 4).
– From the intersection of x = 3 and y = x + 1, we get (3, 4).
Now, evaluate Z = 3x + 4y at the corner points:
– At (0, 0), Z = 3(0) + 4(0) = 0
– At (3, 4), Z = 3(3) + 4(4) = 9 + 16 = 25
The maximum value of Z occurs at (3, 4) and is 25.
Check this for more solutions:
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If the minimum value of an objective function Z = ax + by occurs at two points (3, 4) and (4, 3), then
Given the objective function Z = ax + by, the minimum value occurs at two points (3, 4) and (4, 3). 1. At point (3, 4), the objective function is Z = a(3) + b(4) = 3a + 4b. 2. At point (4, 3), the objective function is Z = a(4) + b(3) = 4a + 3b. Since both points give the minimum value of Z, the objRead more
Given the objective function Z = ax + by, the minimum value occurs at two points (3, 4) and (4, 3).
1. At point (3, 4), the objective function is Z = a(3) + b(4) = 3a + 4b.
2. At point (4, 3), the objective function is Z = a(4) + b(3) = 4a + 3b.
Since both points give the minimum value of Z, the objective function at these points must be equal:
3a + 4b = 4a + 3b
Solving for a and b:
3a + 4b – 4a – 3b = 0
-a + b = 0 ⟹ a = b
Thus, the correct condition is a = b.
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The maximum value of Z = 3x + 4y subject to the constraints x ≥ 0, y ≥ 0 and x + y ≤ 1 is
To solve for the maximum value of Z = 3x + 4y subject to the constraints: 1. x ≥ 0 2. y ≥ 0 3. x + y ≤ 1 We plot the constraints first and determine the feasible region. The feasible region is bounded by the points: - (0, 0) where x = 0 and y = 0 - (1, 0) where x + y = 1 and y = 0 - (0, 1) where x +Read more
To solve for the maximum value of Z = 3x + 4y subject to the constraints:
1. x ≥ 0
2. y ≥ 0
3. x + y ≤ 1
We plot the constraints first and determine the feasible region.
The feasible region is bounded by the points:
– (0, 0) where x = 0 and y = 0
– (1, 0) where x + y = 1 and y = 0
– (0, 1) where x + y = 1 and x = 0
Then we calculate Z = 3x + 4y at all the corner points:
– At (0, 0), Z = 3(0) + 4(0) = 0
– At (1, 0), Z = 3(1) + 4(0) = 3
– At (0, 1), Z = 3(0) + 4(1) = 4
The maximum value for Z is 4 and it occurs at (0, 1).
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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.
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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.
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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).
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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).
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