Determine the conditional extreme for $u = x^2 + y^2 + z^2$ for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$. Determine the conditional extreme for $u = x^2 + y^2 + z^2$ for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$.
Using Lagrange's multipliers we get:
$$<2x, 2y, 2z> = \lambda<\frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2}>$$
concluding $\lambda = \frac{1}{a^2} = \frac{1}{b^2} = \frac{1}{c^2} \rightarrow a^2=b^2=c^2$. If so, then this provides the condition that :
$$<x,y,z> = \frac{\lambda}{a^2}<{x}, {y}, z>$$
Then either $<x,y,z> = <0,0,0>$ or $\lambda = a^2$ (or both).  I can't seem to conclude an extrema condition from this. What am I missing?

$a,b,c$ are all constants so you cannot conclude that they are all equal. What you can conclude from your four equations is that either

*

*$\lambda = a^2, y = z = 0$, $x = \pm a$

*$\lambda = b^2, x = z = 0$, $y = \pm b$

*$\lambda = c^2, x = y = 0$, $z = \pm c$
Then, you can take the maximum of $a,b,c$ and get your extrema conditions.
