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What I actually have to do is find the point on the surface which is closest to the point $P$, both of which are given below.

Function $z(x,y)=x^2+2y^2$

Point $P = (1,-1,1)$

This is what I have tried:

Consider a point $Q = (a,b,c)$ on this surface. Let this be the point that is closest to the point $P$. Since $Q \in z(x,y)$, we have that $c=a^2+2b^2$, so the point is $Q = (a,b,a^2+2b^2 )$. The vector $PQ$ is perpendicular to the surface at the point $(a,b,c)$.

$PQ=(a−1,b+1,a^2+2b^2−1)$

The distance:

$$ \begin{align*} D&=\sqrt{(a−1)^2+(b+1)^2+(a^2+2b^2−1)^2} \\ &=\sqrt{a^2−2a+1+b^2+2b+1+a^4+2a^2 b^2−a^2+2a^2 b^2+4b^4−2b^2−a^2−2b^2+1} \\ &=\sqrt{3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4} \end{align*} $$

Minimize the distance by setting partial derivatives equal to 0:

$$ \begin{align*} D_a&=\frac{8b^2 a−2−2a+4a^3}{\sqrt{2(3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4}} \\ &=\frac{4a^3+(8b^2−2)a−2}{\sqrt{2(3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4}} \\ &\Rightarrow 0=4a^3+(8b^2−2)a−2 \end{align*} $$ $$ \begin{align*} D_b&=\frac{16b^3+(8a^2−6)b+2}{\sqrt{3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4}} \\ &\Rightarrow 0=16b^3+(8a^2−6)b+2 \end{align*} $$ Basically this is where I've gotten to based on some of the suggestions I have seen for similar questions. I saw one method which looked nice here: (http://www.physicsforums.com/showthread.php?t=340772) but they lose me when they get the normal vector, which I have no idea where it came from. I have the actual answer (0.728, -0.573, 1.187) but cannot for the life in me figure out how to get there. I'm looking for errors in my arithmetic or another method if somebody has one (note: I do not know Lagrange Multipliers, which I have seen applied in some solutions, and they are not part of my curriculum, so it would not be useful to solve them this way)

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    $\begingroup$ You can minimize the square of the distance, rather than the distance itself. This will make the algebra a lot simpler. Your approach is basically correct, so it should work. $\endgroup$ – bubba Feb 28 '13 at 2:46
  • $\begingroup$ The surface normal vector is the cross product of the partial derivatives $\tfrac{\partial z}{\partial x}$ and $\tfrac{\partial z}{\partial y}$. But, if you use this, I suspect that you'll end up with the same equations you already have. $\endgroup$ – bubba Feb 28 '13 at 2:54
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    $\begingroup$ To check that your algebra is correct, see if $a = 0.728$, $b = -0.573$ is a solution to your two equations. If it is, you're done. $\endgroup$ – bubba Feb 28 '13 at 2:57
  • $\begingroup$ I got the same equations minimizing the square of the distance. I have just always had trouble trying to factor and minimize polynomials with degree greater than 2. I subbed the point given in the question in and it satisfies it, so I know my algebra up to here is right. That should be enough to get the marks. Thanks for the hand everybody. $\endgroup$ – Yak Attack Mar 3 '13 at 21:37

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