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).
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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