Lagrange Multipliers with two constraints The problem is to find the maximum value of $ \ f(x,y,z) \ = \ x+y+z \ $ subject to the two constraints $ \ g(x,y,z) \ = \ x^2+y^2+z^2 \ = \ 9 \ $ and $ \ h(x,y,z) \ = \ \frac{1}{4}x^2+\frac{1}{4}y^2+4z^2 \ = \ 9 \ $. 
I got these equations:
$$1 \ = \ 2x \ λ \ + \ \frac{1}{2}  x \ μ \ \ , \ \ 1 \ = \ 2y \ λ + \frac{1}{2}y \ μ \ \ , \ \
1 \ = \ 2z \ λ + 8z \ μ \ \ . $$
And from here, I'm not sure what I can solve for or simplify.
 A: Applying the indicated "Lagrange equations", we can subtract the second from the first to produce
$$ ( \ 2 \lambda \ + \ \frac{1}{2} \ \mu \ ) \ ( \ x \ - \ y \ ) \ = \ 0  \ \ . $$
This makes it a bit more evident that one's intuition about the symmetry of the functions is helpful.  As Ted Shifrin already notes, one result from this is that $ \ x \ = \ y \ $ .  Inserting this into the constraint equations, we obtain
$$ 2 \ x^2 \ + \ z^2 \ = \ 9 \ \ , \ \ \frac{1}{2} \ x^2 \ + \ 4 \ z^2 \ = \ 9 \ \ \Rightarrow \ \ \frac{3}{2} \ x^2 \ = \ 3 \ z^2 $$
$$ x^2 \ = \ 2 \ z^2 \ \ \Rightarrow \ \ x \ = \ y \ = \ \pm \sqrt{2} \ z $$
$$ 5 \ z^2 \ = \ 9 \ \ \Rightarrow \ \ z \ = \ \pm \frac{3 \sqrt{5}}{5} \ \ , \ \ x \ = \ y \ = \ \pm \frac{3 \sqrt{10}}{5} \ \ . $$
Hence, the maximum value of the function $ \ f(x, \ y, \ z) \ = \ x + y + z \ $ is found in the first octant as $ \ \frac{3 \sqrt{5}}{5} \ + \ \frac{3 \sqrt{5}}{5} \ + \ \frac{3 \sqrt{10}}{5} \ = \ \frac{3 \sqrt{5}}{5} \ ( \ 2 \sqrt{2} \ + \ 1 \ ) \ $ .  Because the constraint surfaces are symmetric about the origin, and the function to be extremized is anti-symmetric about the origin, we would expect that the minimum value for the function is $ \ -\frac{3 \sqrt{5}}{5} \ ( \ 2 \sqrt{2} \ + \ 1 \ ) \ $ , found in the opposing octant at the point on the constraint surfaces intersected by a line through the origin from the point we just found.
The other "result" from the first equation is $ \ \lambda \ = \ -4 \ \mu \ $ .  Inserting this into the third Lagrange equation gives us $ \ 2 \ z \ \lambda \ = \ 8 \ ( \ -4 \ \lambda \ ) z \ = \ 1 $ $ \ \Rightarrow \ \lambda \ = \ -\frac{1}{30 \ z} \ \ , \ \ \mu \ = \ \frac{2}{15 \ z} \ $ .  This, however, doesn't appear to take us any further.
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the constraint sphere and constraint oblate spheroid, with one intersection circle marked in yellow (the other is located symmetrically "below the equator"); the minimum of the function to be extremized lies on the far side of the surfaces depicted, on the "lower" circle [the "beveling" near the equator of the oblate spheroid is an artifact of the grapher]
If we look at the geometric situation, the two constraint surfaces are a sphere and a (highly) oblate spheroid, which intersect at two circles in planes parallel to the $ \ xy-$ plane and disposed symmetrically about it (producing the rather saturnian -- or perhaps flying-saucerish -- configuration in the diagram above).  We can locate those planes by solving for the intersections of the spheroids:
$$ z^2 \ = \ 9 \ - \ x^2 \ - \ y^2 \ = \ \frac{1}{4} \ (9 \ - \ \frac{1}{4} \ x^2 \ - \ \frac{1}{4} \ y^2) \ $$
$$ 144 \ - \ 16 \ x^2 \ - \ 16 \ y^2 \ = \ 36 \ - \ x^2 \ - \ y^2 \ \ \Rightarrow \ \ 15 \ (x^2 \ + \ y^2) \ = \ 108  $$
$$ \ \ \Rightarrow \ \ x^2 \ + \ y^2 \ = \ \frac{108}{15} \ = \ \frac{36}{5} \ \ \Rightarrow \ \ z^2 \ = \ 9 \ - \ \frac{36}{5} \ = \ \frac{9}{5} \ \ . $$
This yields the values we found above.  We can "reduce" the function to that of a single variable, $ \ f(x) \ = \ x \ \pm \ \left( \ \frac{36}{5} \ - \ x^2 \ \right)^{1/2} \ \pm \frac{3 \sqrt{5}}{5} \ $ to locate extrema thus:
$$ 1 \ \pm \ \frac{1}{2} \ \left( \ \frac{36}{5} \ - \ x^2 \ \right)^{-1/2}  \cdot \ ( -2x ) \ = \ 0 \ \ \Rightarrow \ \ \mp \frac{x}{\sqrt{\frac{36}{5} \ - \ x^2}} \ = \ 1 $$
$$ \frac{36}{5} \ - \ x^2 \ = \ x^2 \ \ \Rightarrow \ \ x^2 \ = \ \frac{18}{5} \ \ , $$
from which follow the results described above.
