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if $f(x,y,z)=x+y-z$ and constraint is sphere $g(x,y,z)=x^2+y^2+z^2=81$. I have to find the absolute maximum and minimum. And I have to explain how both an absolute maximum and minimum must be attained.

my solution: I obtained critical points $(-3\sqrt{3} ,-3\sqrt{3},3\sqrt{3}) and (3\sqrt{3} ,3\sqrt{3},-3\sqrt{3})$ by solving the system $\nabla f= \lambda \nabla g$ and constraint equation.

Then I used the equation $L(l;x,y,z) =x+y-z-l(x^2+y^2+z^2-81)$ and found the matrix $\begin{bmatrix} 0 & -2x & -2y & -2z\\ -2x & -2l & 0 & 0\\ -2y & 0 & -2l & 0\\ -2z & 0 & 0 & -2l \end{bmatrix}$.

I plugged in the critical points, found the sequence of determinants and found that $(-3\sqrt{3} ,-3\sqrt{3},3\sqrt{3})$ is local minimum and $(3\sqrt{3} ,3\sqrt{3},-3\sqrt{3})$ is local maximum.

How do I say that they are absolute maximum and minimum, as well as that they both must be attained?

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  • $\begingroup$ @dxiv I corrected the error about wrong critical points, but can you please explain to me about the second part of the question? $\endgroup$ – kronos Jul 6 '18 at 23:46
  • $\begingroup$ No matter how you solve the first part, once you do solve it the second part should become straightforward to solve in a similar way. Just note that the global minimum of $x+y-z$ is the global maximum of $-x-y+z\,$. $\endgroup$ – dxiv Jul 6 '18 at 23:50
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Hint (for the 1st part): using the RMS-AM inequality, and also that $|a| \ge a$ and $|a| \ge -a$:

$$ 3 \sqrt{3} = \sqrt{\frac{81}{3}} = \sqrt{\frac{x^2+y^2+z^2}{3}} \ge \frac{|x|+|y|+|z|}{3} \ge \frac{x+y-z}{3} \\ \iff\quad x+y -z \le 9 \sqrt{3} $$

Equality is attained iff $|x|=|y|=|z|=3 \sqrt{3}$ and $x,y \ge 0, z \le 0$, which gives $x=y=3 \sqrt{3}$ and $z = -3 \sqrt{3}$, so the upper bound of $9 \sqrt{3}$ is indeed a global maximum.

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By the Cauchy-Schwarz inequality one has $f^2 \le (1^2+1^2+(-1)^2)(x^2+y^2+z^2)=243$ and equality only occurs when $(x,y,z)$ is a multiple of $(1,1,-1)$, so the extrema are at $(t,t,-t)$ with $3t^2=81$ and therefore $t=\pm 3\sqrt 3$.

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