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Let $f(x,y)=\frac {x^2}{2}+\frac {y^2}{4}$ on $\{(x,y)|x^2-y^2=2\}$. Find the absolute maximum and minimum if they exist.

I approached this problem using lagrange multiplier, with $g(x,y)=x^2-y^2-2$.

$f_x=x, f_y=y$ and $g_x=2x, g_y=-2y$. Setting these equal, I got $1=2\lambda$ and $-1=2\lambda$, which has no solution.

Can someone please tell me where I did wrong?

Thanks!

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    $\begingroup$ Minimize a one variable function by substituting $y^2$ by $x^2-2$. $\endgroup$ – Pedro Tamaroff Jan 27 '14 at 1:15
  • $\begingroup$ I would rather propose $\dfrac{x^2}{2} = 1+\dfrac{y^2}{2}$. But it's the same really. $\endgroup$ – user88595 Jan 27 '14 at 1:17
  • $\begingroup$ Isn't $ \ f_y \ = \frac{y}{2} \ $ ? $\endgroup$ – colormegone Jan 27 '14 at 4:41
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To answer the actual question, you should be getting three equalities:

$$ \left\{ \begin{array}{rl} x-2\lambda x & = 0, \\ y/2+2\lambda y & =0, \\ x^2-y^2 & =2 \end{array} \right.$$

So you know that $y=0$ and hence that $x=\sqrt{2}$.

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    $\begingroup$ Yes, in short, OP should not have simply divided $ \ \frac{y}{2} \ = \ - 2y \cdot \lambda \ , $ as that "discards" the solution $ \ y = 0 \ . $ $\endgroup$ – colormegone Jan 27 '14 at 4:47

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