QuestionFind the extrema of $f$ subject to the stated constraint. $f(x, y, z)=x-y+z$, subject to $x^{2}+y^{2}+z^{2}=7$ maximum $\quad(x, y, z)=(\square)$ minimum $\quad(x, y, z)=(\square)$
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Studdy Solution
Evaluate $f$ at the critical points:
For $\lambda = \sqrt{\frac{3}{28}}$:
$$f(x, y, z) = -\frac{1}{2\sqrt{\frac{3}{28}}} - \frac{1}{2\sqrt{\frac{3}{28}}} - \frac{1}{2\sqrt{\frac{3}{28}}} = -\frac{3}{2\sqrt{\frac{3}{28}}}$$
For $\lambda = -\sqrt{\frac{3}{28}}$:
$$f(x, y, z) = \frac{1}{2\sqrt{\frac{3}{28}}} + \frac{1}{2\sqrt{\frac{3}{28}}} + \frac{1}{2\sqrt{\frac{3}{28}}} = \frac{3}{2\sqrt{\frac{3}{28}}}$$
The maximum value is $\frac{3}{2\sqrt{\frac{3}{28}}}$ at $(x, y, z) = \left(\frac{1}{2\sqrt{\frac{3}{28}}}, -\frac{1}{2\sqrt{\frac{3}{28}}}, \frac{1}{2\sqrt{\frac{3}{28}}}\right)$.
The minimum value is $-\frac{3}{2\sqrt{\frac{3}{28}}}$ at $(x, y, z) = \left(-\frac{1}{2\sqrt{\frac{3}{28}}}, \frac{1}{2\sqrt{\frac{3}{28}}}, -\frac{1}{2\sqrt{\frac{3}{28}}}\right)$.