Find extreme values of $f$ subject to both constraints. $f=z; g=x^2+y^2=z^2; h=x+y+z=24$
$x = y\\ 2x+z = 24\\ 2x^2 = z^2\\ x = \pm \frac {\sqrt 2}{2} z$
$(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)\\ $
Related videos on Youtube
Nathan
Updated on December 10, 2020Comments

Nathan almost 3 years
$f=z \\ g=x^2+y^2=z^2\\ h=x+y+z=24$
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.

Guacho Perez almost 7 yearsYou already know that $x=y$ and that $z=1y$, you could try plugging these into $h$ to solve for $y$

Doug M almost 7 years$f_z: 1 = 2\lambda z + \mu \implies z=\frac{\mu}{2 \lambda}\frac {1}{2\lambda}$

Guacho Perez almost 7 years@DougM you are right

Ahmed S. Attaalla almost 7 yearsThese problems can be simplified with symmetry. Note that we can switch $x$ and $y$ in $g$ and end up with the same thing. Also the same with $x$ and $y$ in $h$. So we can in conclude that the extreme value occurs somewhere along $x=y$.
