# Find extreme values of $f$ subject to both constraints. $f=z; g=x^2+y^2=z^2; h=x+y+z=24$

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$x = y\\ 2x+z = 24\\ 2x^2 = z^2\\ x = \pm \frac {\sqrt 2}{2} z$

$(1+\frac {\sqrt 2}{2}) z = 24, (1-\frac {\sqrt 2}{2}) z = 24\\ z = \frac {48}{2+\sqrt2}, z = \frac {48}{2-\sqrt2}\\ z = \frac 24(2-\sqrt2), z = 24(2+\sqrt2)\\$

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### Nathan

Updated on December 10, 2020

• 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 years
You already know that $x=y$ and that $z=1-y$, 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 years
These 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 $x=y$.