91Ó°ÊÓ

First, we define the Lagrangian function L(x, y, z, λ) as follows: $$L(x, y, z, \lambda) = f(x, y, z) - \lambda (g(x, y, z) - 5)$$ where f(x, y, z) = x + y + z and g(x, y, z) = x^2 + y^2 + z^2 - xy. So, the Lagrangian function is: $$L(x, y, z, \lambda) = x + y + z - \lambda (x^2 + y^2 + z^2 - xy - 5)$$

Next, we find the partial derivatives of the Lagrangian function L with respect to x, y, z, and λ: $$\frac{\partial L}{\partial x} = 1 - \lambda (2x - y)$$ $$\frac{\partial L}{\partial y} = 1 - \lambda (2y - x)$$ $$\frac{\partial L}{\partial z} = 1 - 2\lambda z$$ $$\frac{\partial L}{\partial \lambda} = x^2 + y^2 + z^2 - xy - 5$$

Now we need to solve the following system of equations: $$1 - \lambda (2x - y) = 0$$ $$1 - \lambda (2y - x) = 0$$ $$1 - 2\lambda z = 0$$ $$x^2 + y^2 + z^2 - xy - 5 = 0$$ Solving the third equation for λ gives: $$\lambda = \frac{1}{2z}$$ Substitute this in the first two equations and re-arrange: $$x - \frac{y}{4z} = 0$$ $$y - \frac{x}{4z} = 0$$ Since x and y cannot be zero simultaneously, z can't be zero. Thus, we can safely eliminate z terms by multiplying the first equation by y and the second equation by x, and then add both equations: $$x^2 + y^2 = \frac{xy}{2}$$ Combine this equation with the constraint equation: $$x^2 + y^2 + z^2 - xy - 5 = 0$$ $$\frac{3xy}{2} + z^2 - 5 = 0$$ Now, substitute z from the λ equation: $$\frac{3xy}{2} + \frac{1}{4z^2} - 5 = 0$$ Solve this equation for x and y, and then use the constraint to solve for z. After solving, we get two points: $$(x, y, z) = \left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}},\frac{4}{\sqrt{2}}\right) \text{ and } \left(-\frac{3}{\sqrt{2}},-\frac{3}{\sqrt{2}},\frac{4}{\sqrt{2}}\right)$$

Now, we substitute these points back into the function f(x, y, z) to find the maxima and minima: At \((x, y, z) = \left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}},\frac{4}{\sqrt{2}}\right)\) : $$f(x, y, z) = \frac{3}{\sqrt{2}}+\frac{3}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{10}{\sqrt{2}}$$ At \((x, y, z) = \left(-\frac{3}{\sqrt{2}},-\frac{3}{\sqrt{2}},\frac{4}{\sqrt{2}}\right)\) : $$f(x, y, z) = -\frac{3}{\sqrt{2}}-\frac{3}{\sqrt{2}}+\frac{4}{\sqrt{2}}=-\frac{2}{\sqrt{2}}$$ Thus, we find that the minimum value of f(x, y, z) is \(-\frac{2}{\sqrt{2}}\) and the maximum value is \(\frac{10}{\sqrt{2}}\).