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Question

# Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema:Evaluate f at each of the solution points and select the extreme value subject to the constraints asked for in the exercise.Minimize $f ( x , y , z ) = x ^ { 2 } + y ^ { 2 } + z ^ { 2 }$ subject to the constraints $x ^ { 2 } - x y + y ^ { 2 } - z ^ { 2 } - 1 = 0 \text { and } x ^ { 2 } + y ^ { 2 } - 1 = 0.$

Solution

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All that remains to be done is to evaluate $f(x,y,z)$ at the points we obtained from the previous part of the exercise and select the minimal value. For the evaluations of $f$ we have:

\begin{aligned} & f(-1,0,0) = 1 +0+0=1, \\ & f(0,-1,0) =0+1 +0=1\\ & f(0,1,0) = 0+1+0=1 \\ & f(1,0,0) = 1 +0+0=1 \\ & f\left( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\ & f\left( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\ & f\left( \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\ & f\left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \end{aligned}

From these values we see that the minimal value of $f$ under the given constraints is 1 and is reached at the points $(-1,0,0),(0,-1,0),(0,1,0),(1,0,0)$.

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