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To determine the maximum and minimum values of the function [tex]\( f(x, y, z) = 6x + 6y + 10z \)[/tex] subject to the constraint [tex]\( g(x, y, z) = x^2 + y^2 + z^2 - 43 = 0 \)[/tex] using Lagrange multipliers, we follow these steps:
1. Define the Lagrangian Function:
The Lagrangian function [tex]\( \mathcal{L} \)[/tex] is given by:
[tex]\[ \mathcal{L}(x, y, z, \lambda) = f(x, y, z) + \lambda g(x, y, z) \][/tex]
Substituting our given functions:
[tex]\[ \mathcal{L}(x, y, z, \lambda) = 6x + 6y + 10z + \lambda (x^2 + y^2 + z^2 - 43) \][/tex]
2. Compute the Partial Derivatives:
Compute the partial derivatives of [tex]\( \mathcal{L} \)[/tex] with respect to [tex]\( x \)[/tex], [tex]\( y \)[/tex], [tex]\( z \)[/tex], and [tex]\( \lambda \)[/tex] and set them to zero to find the critical points.
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 6 + 2\lambda x = 0 \quad (1) \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 6 + 2\lambda y = 0 \quad (2) \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial z} = 10 + 2\lambda z = 0 \quad (3) \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 + z^2 - 43 = 0 \quad (4) \][/tex]
3. Solve the System of Equations:
From equations (1), (2), and (3), we express [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] in terms of [tex]\( \lambda \)[/tex]:
[tex]\[ 6 + 2\lambda x = 0 \implies x = -\frac{3}{\lambda} \quad (5) \][/tex]
[tex]\[ 6 + 2\lambda y = 0 \implies y = -\frac{3}{\lambda} \quad (6) \][/tex]
[tex]\[ 10 + 2\lambda z = 0 \implies z = -\frac{5}{\lambda} \quad (7) \][/tex]
Substitute equations (5), (6), and (7) into the constraint equation (4):
[tex]\[ \left(-\frac{3}{\lambda}\right)^2 + \left(-\frac{3}{\lambda}\right)^2 + \left(-\frac{5}{\lambda}\right)^2 = 43 \][/tex]
Simplifying:
[tex]\[ \frac{9}{\lambda^2} + \frac{9}{\lambda^2} + \frac{25}{\lambda^2} = 43 \][/tex]
[tex]\[ \frac{43}{\lambda^2} = 43 \][/tex]
[tex]\[ \lambda^2 = 1 \implies \lambda = \pm 1 \][/tex]
4. Find the corresponding points [tex]\((x, y, z)\)[/tex]:
For [tex]\( \lambda = 1 \)[/tex]:
[tex]\[ x = -\frac{3}{1} = -3, \quad y = -3, \quad z = -5 \][/tex]
For [tex]\( \lambda = -1 \)[/tex]:
[tex]\[ x = -\frac{3}{-1} = 3, \quad y = 3, \quad z = 5 \][/tex]
5. Evaluate [tex]\( f(x, y, z) \)[/tex] at the critical points:
- For [tex]\( (x, y, z) = (-3, -3, -5) \)[/tex]:
[tex]\[ f(-3, -3, -5) = 6(-3) + 6(-3) + 10(-5) = -18 - 18 - 50 = -86 \][/tex]
- For [tex]\( (x, y, z) = (3, 3, 5) \)[/tex]:
[tex]\[ f(3, 3, 5) = 6(3) + 6(3) + 10(5) = 18 + 18 + 50 = 86 \][/tex]
6. Conclusion:
The maximum value of the function [tex]\( f(x, y, z) \)[/tex] subject to the given constraint is [tex]\( 86 \)[/tex], and the minimum value is [tex]\( -86 \)[/tex].
[tex]\[ \text{maximum} = 86 \][/tex]
[tex]\[ \text{minimum} = -86 \][/tex]
1. Define the Lagrangian Function:
The Lagrangian function [tex]\( \mathcal{L} \)[/tex] is given by:
[tex]\[ \mathcal{L}(x, y, z, \lambda) = f(x, y, z) + \lambda g(x, y, z) \][/tex]
Substituting our given functions:
[tex]\[ \mathcal{L}(x, y, z, \lambda) = 6x + 6y + 10z + \lambda (x^2 + y^2 + z^2 - 43) \][/tex]
2. Compute the Partial Derivatives:
Compute the partial derivatives of [tex]\( \mathcal{L} \)[/tex] with respect to [tex]\( x \)[/tex], [tex]\( y \)[/tex], [tex]\( z \)[/tex], and [tex]\( \lambda \)[/tex] and set them to zero to find the critical points.
[tex]\[ \frac{\partial \mathcal{L}}{\partial x} = 6 + 2\lambda x = 0 \quad (1) \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial y} = 6 + 2\lambda y = 0 \quad (2) \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial z} = 10 + 2\lambda z = 0 \quad (3) \][/tex]
[tex]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 + z^2 - 43 = 0 \quad (4) \][/tex]
3. Solve the System of Equations:
From equations (1), (2), and (3), we express [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] in terms of [tex]\( \lambda \)[/tex]:
[tex]\[ 6 + 2\lambda x = 0 \implies x = -\frac{3}{\lambda} \quad (5) \][/tex]
[tex]\[ 6 + 2\lambda y = 0 \implies y = -\frac{3}{\lambda} \quad (6) \][/tex]
[tex]\[ 10 + 2\lambda z = 0 \implies z = -\frac{5}{\lambda} \quad (7) \][/tex]
Substitute equations (5), (6), and (7) into the constraint equation (4):
[tex]\[ \left(-\frac{3}{\lambda}\right)^2 + \left(-\frac{3}{\lambda}\right)^2 + \left(-\frac{5}{\lambda}\right)^2 = 43 \][/tex]
Simplifying:
[tex]\[ \frac{9}{\lambda^2} + \frac{9}{\lambda^2} + \frac{25}{\lambda^2} = 43 \][/tex]
[tex]\[ \frac{43}{\lambda^2} = 43 \][/tex]
[tex]\[ \lambda^2 = 1 \implies \lambda = \pm 1 \][/tex]
4. Find the corresponding points [tex]\((x, y, z)\)[/tex]:
For [tex]\( \lambda = 1 \)[/tex]:
[tex]\[ x = -\frac{3}{1} = -3, \quad y = -3, \quad z = -5 \][/tex]
For [tex]\( \lambda = -1 \)[/tex]:
[tex]\[ x = -\frac{3}{-1} = 3, \quad y = 3, \quad z = 5 \][/tex]
5. Evaluate [tex]\( f(x, y, z) \)[/tex] at the critical points:
- For [tex]\( (x, y, z) = (-3, -3, -5) \)[/tex]:
[tex]\[ f(-3, -3, -5) = 6(-3) + 6(-3) + 10(-5) = -18 - 18 - 50 = -86 \][/tex]
- For [tex]\( (x, y, z) = (3, 3, 5) \)[/tex]:
[tex]\[ f(3, 3, 5) = 6(3) + 6(3) + 10(5) = 18 + 18 + 50 = 86 \][/tex]
6. Conclusion:
The maximum value of the function [tex]\( f(x, y, z) \)[/tex] subject to the given constraint is [tex]\( 86 \)[/tex], and the minimum value is [tex]\( -86 \)[/tex].
[tex]\[ \text{maximum} = 86 \][/tex]
[tex]\[ \text{minimum} = -86 \][/tex]
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