Determine the conditional extreme for $u = x^2 + y^2 + z^2$ for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$.

Question

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?

Answer 1

$a,b,c$ 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.