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I am supposed to find the points (x,y,z) satisfying the condition $x^2+2y^2-z^2-1=0$ that are the closest to origin (0,0,0). So basically, the idea was to find the minima of $$\Lambda(x,y,z,\lambda) = \sqrt{x^2+y^2+z^2}+\lambda(x^2+2y^2-2z^2-1)$$

For that I determined $$\Lambda_x(x,y,z,\lambda) = \frac{x}{\sqrt{x^2+y^2+z^2}} + 2\lambda x$$ $$\Lambda_y(x,y,z,\lambda) = \frac{y}{\sqrt{x^2+y^2+z^2}} + 4\lambda y$$ $$\Lambda_z(x,y,z,\lambda) = \frac{z}{\sqrt{x^2+y^2+z^2}} - 2\lambda z$$

But that would mean for $\lambda \neq 0$ $$\lambda_1 = -\frac{1}{2\sqrt{x^2+y^2+z^2}}$$ $$\lambda_2 = -\frac{1}{4\sqrt{x^2+y^2+z^2}}$$ $$\lambda_3 = \frac{1}{2\sqrt{x^2+y^2+z^2}}$$

Since $\lambda_1 \neq \lambda_2 \neq \lambda_3$ doesn't that mean that this function doesn't have any minima? I'm probably having a huge misunderstanding somewhere, but I just can't figure out where I went wrong, so please help me with this.

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2 Answers

First, it's easier to minimize the square of the distance, rather than the distance. I guess that means want to minimize $f(x,y,z)=x^2+y^2+z^2$ subject to $g(x,y,z) = x^2+2y^2-z^2=1$. I think that means you want to solve the system $\nabla f=\lambda g$ or $$2x = 2\lambda x,$$ $$2y=4\lambda y,$$ $$2z = -2z^2,$$ $$x^2+2y^2-z^2=1.$$

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I don't know, I am using the notation I am used to. But unless the distance happens to be 1, why would I be able to use the square of the distance rather than the distance? EDIT: Actually, that makes sense because I can just take the square root when I have the result... EDIT2: Your signs seem off though. –  Cubic May 30 '12 at 20:42
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The problem with your reasoning is that it assumes that neither $x$, $y$ or $z$ is $0$, but this is false!

It's easier to solve this problem using the square of distance instead. The distance is always positive, so if you minimize the square of distance, you'll minimize the distance too.

We have:

$$\Lambda(x,y,z,\lambda) = x^2+y^2+z^2 + \lambda(x^2+2y^2-z^2-1)$$

Therefore:

\begin{align*} \Lambda_x(x,y,z,\lambda) &= 2x + 2\lambda x &= 0 \\ \Lambda_y(x,y,z,\lambda) &= 2y + 4\lambda y &= 0 \\ \Lambda_z(x,y,z,\lambda) &= 2z - 2\lambda z &= 0 \\ \Lambda_\lambda(x,y,z,\lambda) &= x^2+2y^2-z^2-1 &= 0 \end{align*}

Or:

\begin{align*} (1+\lambda)x &= 0 \\ (1+2\lambda)y &= 0 \\ (1-\lambda)z &= 0 \\ x^2+2y^2-z^2-1 &= 0 \end{align*}

For each of the first three equations use: If $a \cdot b = 0$, then $a = 0$ or $b = 0$. It'll enable you to find $(x, y, z)$ that minimizes the distance.

Also, notice that $x^2+2y^2-z^2-1 = 0$ is a hyperboloid of one sheet. Look at its plot and try to guess the points that minimize the distance. Compare with your answer above.

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Thanks. Will do. –  Cubic May 30 '12 at 21:37
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