a. Use Lagrange Multiplier (LM) to find points on the surface z=x2+y2 closest to the point (0,0,2). b. Consider the two surfaces given by: S1 = x2+y2+z2-4x+1=0 and S2 = y2-x+1. Find the maximum/minimum distance for points in both S1 and S2. (Hint: Use LM)

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Asked on June 7, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-06-07T06:53:21.961Z

Let's tackle Question Three. Question Three a) Use Lagrange Multiplier (LM) to find points on the surface z = x^2 + y^2 closest to the point (0,0,2). Step 1: Define the objective function and the constraint. We want to minimize the distance between a point (x,y,z) on the surface and (0,0,2). It's easier to minimize the square of the distance: f(x,y,z) = (x-0)^2 + (y-0)^2 + (z-2)^2 = x^2 + y^2 + (z-2)^2 The constraint is that the point (x,y,z) lies on the surface z = x^2 + y^2, which can be written as: g(x,y,z) = x^2 + y^2 - z = 0 Step 2: Set up the Lagrange Multiplier equations. We need to solve f = g. f_x = 2x, f_y = 2y, f_z = 2(z-2) g_x = 2x, g_y = 2y, g_z = -1 The system of equations is: 1. 2x = (2x) 2. 2y = (2y) 3. 2(z-2) = (-1) 4. x^2 + y^2 - z = 0 (the constraint) Step 3: Solve the system of equations. From (1), 2x(1-) = 0, which implies x=0 or =1. From (2), 2y(1-) = 0, which implies y=0 or =1. Case 1: x=0 and y=0 Substitute into (4): 0^2 + 0^2 - z = 0 z=0. So, the point is (0,0,0). Check with (3): 2(0-2) = - -4 = - =4. This is consistent. Case 2: =1 Substitute into (3): 2(z-2) = -1 2z - 4 = -1 2z = 3 z = (3)/(2). Substitute z = (3)/(2) into (4): x^2 + y^2 - (3)/(2) = 0 x^2 + y^2 = (3)/(2). This gives a set of points (x,y,z) where x^2+y^2 = (3)/(2) and z=(3)/(2). Step 4: Evaluate the objective function at the candidate points. For (0,0,0): f(0,0,0) = 0^2 + 0^2 + (0-2)^2 = 4. The distance is sqrt(4) = 2. For points where x^2+y^2 = (3)/(2) and z=(3)/(2): f(x,y,z) = (x^2+y^2) + (z-2)^2 = (3)/(2) + ((3)/(2)-2)^2 = (3)/(2) + (-(1)/(2))^2 = (3)/(2) + (1)/(4) = (6)/(4) + (1)/(4) = (7)/(4). The distance is sqrt((7)/(4)) = sqrt(7)2. Comparing the distances, 2 ≈ 1.414 and sqrt(7)2 ≈ (2.646)/(2) ≈ 1.323. The minimum distance is sqrt(7)2. The points on the surface closest to (0,0,2) are all points (x,y,z) such that x^2+y^2 = (3)/(2) and z=(3)/(2). b) Consider the two surfaces given by S_1 = x^2 + y^2 + z^2 - 4x + 1 = 0 and S_2 = y^2 - x + 1 = 0. Find the maximum/minimum distance for points in both S_1 and S_2. (Hint: Use LM) This asks for the maximum and minimum distance from the origin (0,0,0) to points (x,y,z) that lie on the intersection of S_1 and S_2. Step 1: Define the objective function and the constraints. We want to minimize/maximize the square of the distance from the origin: f(x,y,z) = x^2 + y^2 + z^2 The two constraints are: g_1(x,y,z) = x^2 + y^2 + z^2 - 4x + 1 = 0 g_2(x,y,z) = y^2 - x + 1 = 0 Step 2: Set up the Lagrange Multiplier equations. We need to solve f = _1 g_1 + _2 g_2. f_x = 2x, f_y = 2y, f_z = 2z g_1x = 2x-4, g_1y = 2y, g_1z = 2z g_2x = -1, g_2y = 2y, g_2z = 0 The system of equations is: 1. 2x = _1(2x-4) + _2(-1) 2. 2y = _1(2y) + _2(2y) 3. 2z = _1(2z) 4. x^2 + y^2 + z^2 - 4x + 1 = 0 5. y^2 - x + 1 = 0 Step 3: Solve the system of equations. From (3): 2z = 2_1 z 2z(1-_1) = 0. This implies z=0 or _1=1. Case 1: z=0 Substitute z=0 into (4): x^2 + y^2 - 4x + 1 = 0. From (5): y^2 - x + 1 = 0 y^2 = x - 1. Substitute y^2 = x-1 into x^2 + y^2 - 4x + 1 = 0: x^2 + (x-1) - 4x + 1 = 0 x^2 - 3x = 0 x(x-3) = 0 This gives x=0 or x=3. If x=0, then y^2 = 0-1 = -1, which has no real solutions for y. If x=3, then y^2 = 3-1 = 2 y = ±sqrt(2). So, two candidate points are (3, sqrt(2), 0) and (3, -sqrt(2), 0). Case 2: _1=1 Substitute _1=1 into (2): 2y = 1(2y) + _2(2y) 2y = 2y + 2_2 y 2_2 y = 0. This implies _2=0 or y=0. If _2=0: Substitute _1=1, _2=0 into (1): 2x = 1(2x-4) + 0(-1) 2x = 2x-4 0 = -4. This is a contradiction, so _2 0. Therefore, we must have y=0. Substitute y=0 into (5): 0^2 - x + 1 = 0 x=1. Substitute x=1, y=0 into (4): 1^2 + 0^2 + z^2 - 4(1) + 1 = 0 1 + z^2 - 4 + 1 = 0 z^2 - 2 = 0 z = ±sqrt(2). So, two more candidate points are (1, 0, sqrt(2)) and (1, 0, -sqrt(2)). Step 4: Evaluate the objective function f(x,y,z) = x^2 + y^2 + z^2 at the candidate points. For (3, sqrt(2), 0): f(3, sqrt(2), 0) = 3^2 + (sqrt(2))^2 + 0^2 = 9 + 2 + 0 = 11. The distance is sqrt(11). For (3, -sqrt(2), 0): f(3, -sqrt(2), 0) = 3^2 + (-sqrt(2))^2 + 0^2 = 9 + 2 + 0 = 11. The distance is sqrt(11). For (1, 0, sqrt(2)): f(1, 0, sqrt(2)) = 1^2 + 0^2 + (sqrt(2))^2 = 1 + 0 + 2 = 3. The distance is sqrt(3). For (1, 0, -sqrt(2)): f(1, 0, -sqrt(2)) = 1^2 + 0^2 + (-sqrt(2))^2 = 1 + 0 + 2 = 3. The distance is sqrt(3). Step 5: Determine the maximum and minimum distances. Comparing the distances: sqrt(11) ≈ 3.317 and sqrt(3) ≈ 1.732. The maximum distance is sqrt(11). The minimum distance is sqrt(3). The maximum distance is sqrt(11) and the minimum distance is sqrt(3). That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-06-07T06:53:21.961Z

Let's tackle Question Three. Question Three a) Use Lagrange Multiplier (LM) to find points on the surface z = x^2 + y^2 closest to the point (0,0,2). Step 1: Define the objective function and the constraint. We want to minimize the distance between a point (x,y,z) on the surface and (0,0,2). It's easier to minimize the square of the distance: f(x,y,z) = (x-0)^2 + (y-0)^2 + (z-2)^2 = x^2 + y^2 + (z-2)^2 The constraint is that the point (x,y,z) lies on the surface z = x^2 + y^2, which can be written as: g(x,y,z) = x^2 + y^2 - z = 0 Step 2: Set up the Lagrange Multiplier equations. We need to solve f = g. f_x = 2x, f_y = 2y, f_z = 2(z-2) g_x = 2x, g_y = 2y, g_z = -1 The system of equations is: 1. 2x = (2x) 2. 2y = (2y) 3. 2(z-2) = (-1) 4. x^2 + y^2 - z = 0 (the constraint) Step 3: Solve the system of equations. From (1), 2x(1-) = 0, which implies x=0 or =1. From (2), 2y(1-) = 0, which implies y=0 or =1. Case 1: x=0 and y=0 Substitute into (4): 0^2 + 0^2 - z = 0 z=0. So, the point is (0,0,0). Check with (3): 2(0-2) = - -4 = - =4. This is consistent. Case 2: =1 Substitute into (3): 2(z-2) = -1 2z - 4 = -1 2z = 3 z = (3)/(2). Substitute z = (3)/(2) into (4): x^2 + y^2 - (3)/(2) = 0 x^2 + y^2 = (3)/(2). This gives a set of points (x,y,z) where x^2+y^2 = (3)/(2) and z=(3)/(2). Step 4: Evaluate the objective function at the candidate points. For (0,0,0): f(0,0,0) = 0^2 + 0^2 + (0-2)^2 = 4. The distance is sqrt(4) = 2. For points where x^2+y^2 = (3)/(2) and z=(3)/(2): f(x,y,z) = (x^2+y^2) + (z-2)^2 = (3)/(2) + ((3)/(2)-2)^2 = (3)/(2) + (-(1)/(2))^2 = (3)/(2) + (1)/(4) = (6)/(4) + (1)/(4) = (7)/(4). The distance is sqrt((7)/(4)) = sqrt(7)2. Comparing the distances, 2 ≈ 1.414 and sqrt(7)2 ≈ (2.646)/(2) ≈ 1.323. The minimum distance is sqrt(7)2. The points on the surface closest to (0,0,2) are all points (x,y,z) such that x^2+y^2 = (3)/(2) and z=(3)/(2). b) Consider the two surfaces given by S_1 = x^2 + y^2 + z^2 - 4x + 1 = 0 and S_2 = y^2 - x + 1 = 0. Find the maximum/minimum distance for points in both S_1 and S_2. (Hint: Use LM) This asks for the maximum and minimum distance from the origin (0,0,0) to points (x,y,z) that lie on the intersection of S_1 and S_2. Step 1: Define the objective function and the constraints. We want to minimize/maximize the square of the distance from the origin: f(x,y,z) = x^2 + y^2 + z^2 The two constraints are: g_1(x,y,z) = x^2 + y^2 + z^2 - 4x + 1 = 0 g_2(x,y,z) = y^2 - x + 1 = 0 Step 2: Set up the Lagrange Multiplier equations. We need to solve f = _1 g_1 + _2 g_2. f_x = 2x, f_y = 2y, f_z = 2z g_1x = 2x-4, g_1y = 2y, g_1z = 2z g_2x = -1, g_2y = 2y, g_2z = 0 The system of equations is: 1. 2x = _1(2x-4) + _2(-1) 2. 2y = _1(2y) + _2(2y) 3. 2z = _1(2z) 4. x^2 + y^2 + z^2 - 4x + 1 = 0 5. y^2 - x + 1 = 0 Step 3: Solve the system of equations. From (3): 2z = 2_1 z 2z(1-_1) = 0. This implies z=0 or _1=1. Case 1: z=0 Substitute z=0 into (4): x^2 + y^2 - 4x + 1 = 0. From (5): y^2 - x + 1 = 0 y^2 = x - 1. Substitute y^2 = x-1 into x^2 + y^2 - 4x + 1 = 0: x^2 + (x-1) - 4x + 1 = 0 x^2 - 3x = 0 x(x-3) = 0 This gives x=0 or x=3. If x=0, then y^2 = 0-1 = -1, which has no real solutions for y. If x=3, then y^2 = 3-1 = 2 y = ±sqrt(2). So, two candidate points are (3, sqrt(2), 0) and (3, -sqrt(2), 0). Case 2: _1=1 Substitute _1=1 into (2): 2y = 1(2y) + _2(2y) 2y = 2y + 2_2 y 2_2 y = 0. This implies _2=0 or y=0. If _2=0: Substitute _1=1, _2=0 into (1): 2x = 1(2x-4) + 0(-1) 2x = 2x-4 0 = -4. This is a contradiction, so _2 0. Therefore, we must have y=0. Substitute y=0 into (5): 0^2 - x + 1 = 0 x=1. Substitute x=1, y=0 into (4): 1^2 + 0^2 + z^2 - 4(1) + 1 = 0 1 + z^2 - 4 + 1 = 0 z^2 - 2 = 0 z = ±sqrt(2). So, two more candidate points are (1, 0, sqrt(2)) and (1, 0, -sqrt(2)). Step 4: Evaluate the objective function f(x,y,z) = x^2 + y^2 + z^2 at the candidate points. For (3, sqrt(2), 0): f(3, sqrt(2), 0) = 3^2 + (sqrt(2))^2 + 0^2 = 9 + 2 + 0 = 11. The distance is sqrt(11). For (3, -sqrt(2), 0): f(3, -sqrt(2), 0) = 3^2 + (-sqrt(2))^2 + 0^2 = 9 + 2 + 0 = 11. The distance is sqrt(11). For (1, 0, sqrt(2)): f(1, 0, sqrt(2)) = 1^2 + 0^2 + (sqrt(2))^2 = 1 + 0 + 2 = 3. The distance is sqrt(3). For (1, 0, -sqrt(2)): f(1, 0, -sqrt(2)) = 1^2 + 0^2 + (-sqrt(2))^2 = 1 + 0 + 2 = 3. The distance is sqrt(3). Step 5: Determine the maximum and minimum distances. Comparing the distances: sqrt(11) ≈ 3.317 and sqrt(3) ≈ 1.732. The maximum distance is sqrt(11). The minimum distance is sqrt(3). The maximum distance is sqrt(11) and the minimum distance is sqrt(3). That's 2 down. 3 left today — send the next one.

a. Use Lagrange Multiplier (LM) to find points on the surface z=x2+y2 closest to the point (0,0,2). b. Consider the two surfaces given by: S1 = x2+y2+z2-4x+1=0 and S2 = y2-x+1. Find the maximum/minimum distance for points in both S1 and S2. (Hint: Use LM)

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a. Use Lagrange Multiplier (LM) to find points on the surface z=x2+y2 closest to the point (0,0,2). b. Consider the two surfaces given by: S1 = x2+y2+z2-4x+1=0 and S2 = y2-x+1. Find the maximum/minimum distance for points in both S1 and S2. (Hint: Use LM)

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