To use the method of Lagrange multipliers to find the maximum and minimum values of the function [tex]\( f(x, y, z) = 2x + 2y + z \)[/tex] subject to the constraint [tex]\( g(x, y, z) = x^2 + y^2 + z^2 - 9 = 0 \)[/tex], we follow these steps:
1. Set up the Lagrange function:
Define the Lagrangian function [tex]\( \mathcal{L}(x, y, z, \lambda) = f(x, y, z) - \lambda \cdot g(x, y, z) \)[/tex]. Therefore,
[tex]\[
\mathcal{L}(x, y, z, \lambda) = 2x + 2y + z - \lambda (x^2 + y^2 + z^2 - 9).
\][/tex]
2. Compute the partial derivatives:
We need to find the partial derivatives of [tex]\( \mathcal{L} \)[/tex] with respect to [tex]\( x \)[/tex], [tex]\( y \)[/tex], [tex]\( z \)[/tex], and [tex]\( \lambda \)[/tex]:
[tex]\[
\frac{\partial \mathcal{L}}{\partial x} = 2 - 2 \lambda x = 0 \quad \Rightarrow \quad \lambda x = 1
\][/tex]
[tex]\[
\frac{\partial \mathcal{L}}{\partial y} = 2 - 2 \lambda y = 0 \quad \Rightarrow \quad \lambda y = 1
\][/tex]
[tex]\[
\frac{\partial \mathcal{L}}{\partial z} = 1 - 2 \lambda z = 0 \quad \Rightarrow \quad \lambda z = \frac{1}{2}
\][/tex]
[tex]\[
\frac{\partial \mathcal{L}}{\partial \lambda} = - (x^2 + y^2 + z^2 - 9) = 0 \quad \Rightarrow \quad x^2 + y^2 + z^2 = 9
\][/tex]
3. Solve the system of equations:
From the equations derived from the partial derivatives, we have:
[tex]\[
\lambda x = 1, \quad \lambda y = 1, \quad \lambda z = \frac{1}{2}
\][/tex]
This implies:
[tex]\[
x = \frac{1}{\lambda}, \quad y = \frac{1}{\lambda}, \quad z = \frac{1}{2\lambda}
\][/tex]
Substitute these into the constraint [tex]\( x^2 + y^2 + z^2 = 9 \)[/tex]:
[tex]\[
\left( \frac{1}{\lambda} \right)^2 + \left( \frac{1}{\lambda} \right)^2 + \left( \frac{1}{2\lambda} \right)^2 = 9
\][/tex]
[tex]\[
\frac{1}{\lambda^2} + \frac{1}{\lambda^2} + \frac{1}{4\lambda^2} = 9
\][/tex]
[tex]\[
\frac{2}{\lambda^2} + \frac{1}{4\lambda^2} = 9
\][/tex]
[tex]\[
\frac{8}{4\lambda^2} + \frac{1}{4\lambda^2} = 9
\][/tex]
[tex]\[
\frac{9}{4\lambda^2} = 9
\][/tex]
[tex]\[
\frac{9}{4\lambda^2} = 9
\][/tex]
[tex]\[
\frac{1}{\lambda^2} = 1
\][/tex]
[tex]\[
\lambda^2 = 1 \quad \Rightarrow \quad \lambda = \pm 1
\][/tex]
For [tex]\( \lambda = 1 \)[/tex]:
[tex]\[
x = \frac{1}{1} = 1, \quad y = \frac{1}{1} = 1, \quad z = \frac{1}{2} = \frac{1}{2}
\][/tex]
This gives the point [tex]\( (1, 1, \frac{1}{2}) \)[/tex].
For [tex]\( \lambda = -1 \)[/tex]:
[tex]\[
x = \frac{1}{-1} = -1, \quad y = \frac{1}{-1} = -1, \quad z = \frac{1}{2 \cdot -1} = -\frac{1}{2}
\][/tex]
This gives the point [tex]\( (-1, -1, -\frac{1}{2}) \)[/tex].
4. Evaluate [tex]\( f(x, y, z) \)[/tex] at the critical points:
[tex]\[
f(1, 1, \frac{1}{2}) = 2 \cdot 1 + 2 \cdot 1 + \frac{1}{2} = 4.5
\][/tex]
[tex]\[
f(-1, -1, -\frac{1}{2}) = 2 \cdot (-1) + 2 \cdot (-1) + (-\frac{1}{2}) = -4.5
\][/tex]
So, the maximum value is [tex]\( \boxed{4.5} \)[/tex], and the minimum value is [tex]\( \boxed{-4.5} \)[/tex].
