$f=z \\
g=x^2+y^2=z^2\\
h=x+y+z=24$
I'm having a hard time soloing out $\mu, \lambda, x, y,$ or $z$.
$\triangle(f)=\langle0,0,1\rangle\\
\triangle(g)=\langle2x, 2y, -2z\rangle\\
\triangle(h)=\langle1,1,1\rangle$
$
f_x = 0 = \lambda 2x + \mu 1 \rightarrow x=-\frac{\mu 1}{\lambda 2}\\
f_y = 0 = \lambda 2y + \mu 1 \rightarrow y=-\frac{\mu 1}{\lambda 2}\\
f_z = 1 = -\lambda 2z + \mu 1 \rightarrow z=\frac{\mu 1}{\lambda 2}+1
$
If I plug my variables back into $h$, I get $-\frac{\mu 1}{\lambda 2}=25$, this is where I get stuck. I can't say $x=y=-\frac{\mu 1}{\lambda 2}=25$ because that wouldn't work with $g$. I'm at a loss of what to do next.
For the record, this is for Calc III exam prep, not for homework.
Best Answer
$x = y\\ 2x+z = 24\\ 2x^2 = z^2\\ x = \pm \frac {\sqrt 2}{2} z$
$(1+\frac {\sqrt 2}{2}) z = 24, (1-\frac {\sqrt 2}{2}) z = 24\\ z = \frac {48}{2+\sqrt2}, z = \frac {48}{2-\sqrt2}\\ z = \frac 24(2-\sqrt2), z = 24(2+\sqrt2)\\ $