Use Lagrange multiplier to find the maximum and minimum values of the function f(x,y,z) subject to the given constraint. f(x,y,z)=xyz , x^2+y^2+z^2=12.

To find the maximum and minimum values of the function \( f(x, y, z) = xyz \) subject to the constraint \( x^2 + y^2 + z^2 = 12 \), we will use the method of Lagrange multipliers. The method involves introducing a Lagrange multiplier \( \lambda \) and solving the system of equations given by the gradients of the function and the constraint.

1. Define the Lagrangian function:
\[ \mathcal{L}(x, y, z, \lambda) = xyz + \lambda (12 - x^2 - y^2 - z^2) \]

2. Compute the partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), \( z \), and \( \lambda \), and set them equal to zero:
\[ \frac{\partial \mathcal{L}}{\partial x} = yz - 2\lambda x = 0 \]
\[ \frac{\partial \mathcal{L}}{\partial y} = xz - 2\lambda y = 0 \]
\[ \frac{\partial \mathcal{L}}{\partial z} = xy - 2\lambda z = 0 \]
\[ \frac{\partial \mathcal{L}}{\partial \lambda} = 12 - x^2 - y^2 - z^2 = 0 \]

3. Solve the system of equations:

From \( yz - 2\lambda x = 0 \):
\[ yz = 2\lambda x \]

From \( xz - 2\lambda y = 0 \):
\[ xz = 2\lambda y \]

From \( xy - 2\lambda z = 0 \):
\[ xy = 2\lambda z \]

From the constraint:
\[ x^2 + y^2 + z^2 = 12 \]

4. Express \( \lambda \) in terms of \( x, y, z \):

Dividing the first two equations:
\[ \frac{yz}{xz} = \frac{2\lambda x}{2\lambda y} \Rightarrow \frac{y}{x} = \frac{x}{y} \Rightarrow y^2 = x^2 \Rightarrow y = \pm x \]

Dividing the first and third equations:
\[ \frac{yz}{xy} = \frac{2\lambda x}{2\lambda z} \Rightarrow \frac{z}{x} = \frac{x}{z} \Rightarrow z^2 = x^2 \Rightarrow z = \pm x \]

Considering \( y = x \) and \( z = x \), we get:
\[ x^2 + x^2 + x^2 = 12 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \]

Thus, we have the points \((2, 2, 2)\) and \((-2, -2, -2)\).

Considering \( y = -x \) and \( z = x \), we get:
\[ x^2 + (-x)^2 + x^2 = 12 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \]

Thus, we have the points \((2, -2, 2)\) and \((-2, 2, -2)\).

Considering \( y = x \) and \( z = -x \), we get:
\[ x^2 + x^2 + (-x)^2 = 12 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \]

Thus, we have the points \((2, 2, -2)\) and \((-2, -2, 2)\).

5. Evaluate \( f(x, y, z) = xyz \) at these points:

For \((2, 2, 2)\):
\[ f(2, 2, 2) = 2 \cdot 2 \cdot 2 = 8 \]

For \((-2, -2, -2)\):
\[ f(-2, -2, -2) = (-2) \cdot (-2) \cdot (-2) = -8 \]

For \((2, -2, 2)\):
\[ f(2, -2, 2) = 2 \cdot (-2) \cdot 2 = -8 \]

For \((-2, 2, -2)\):
\[ f(-2, 2, -2) = (-2) \cdot 2 \cdot (-2) = 8 \]

For \((2, 2, -2)\):
\[ f(2, 2, -2) = 2 \cdot 2 \cdot (-2) = -8 \]

For \((-2, -2, 2)\):
\[ f(-2, -2, 2) = (-2) \cdot (-2) \cdot 2 = 8 \]

6. Conclusion:

The maximum value of \( f(x, y, z) = xyz \) is \( 8 \), and the minimum value is \( -8 \).

Do you want more details or have any questions?

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Tip: When working with constraints in optimization problems, always check the boundary conditions if they are within the domain of the problem.

Learn how to use Lagrange multipliers to find the maximum and minimum values of the function f(x, y, z) = xyz subject to the constraint x^2 + y^2 + z^2 = 12. Detailed step-by-step solution with evaluations at critical points.

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