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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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If a + ib = c + id, then
Choice (d) is correct. Given, a + ib = c +id ⇒ a = c and b = d ⇒ a² + b² = c² + d² This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding. For more please visit here: https://www.tiwariacadRead more
Choice (d) is correct.
Given, a + ib = c +id
⇒ a = c and b = d
⇒ a² + b² = c² + d²
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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The complex number z = x + iy satisfies the condition ∣ z + 1 ∣ = 1, then z lies on
Choices (d) is correct. Given ∣ z + 1 ∣ = 1 ⇒ ∣ z -(- 1 + 0i ∣ = 1 ⇒ The distance of the point z from the point (-1, 0) is constant and is equal to 1. Thus, z lies on a circle with centre (-1, 0) and radius 1. This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex NRead more
Choices (d) is correct.
Given ∣ z + 1 ∣ = 1 ⇒ ∣ z -(- 1 + 0i ∣ = 1
⇒ The distance of the point z from the point (-1, 0) is constant and is equal to 1.
Thus, z lies on a circle with centre (-1, 0) and radius 1.
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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What does ∣ z ∣ = 4 represent on a plane where z = 4 represent on a plane where z = x + iy, a complex number?
Choice (c) is correct. ∣ z ∣ = 4 ⇒ ∣ x + iy ∣ = 4 ⇒ √x² + y² = 4 ⇒ x² + y² = 16 This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding. For more please visit here: https://www.tiwariacademRead more
Choice (c) is correct.
∣ z ∣ = 4 ⇒ ∣ x + iy ∣ = 4
⇒ √x² + y² = 4 ⇒ x² + y² = 16
This question related to Chapter 4 maths Class 11th NCERT. From the Chapter 4: Complex Numbers and Quadratic Equations. Give answer according to your understanding.
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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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