<s>[INST] Insights needed for the following Lagrange Multipler problem Find the point of the paraboloid $z = \frac{x^2}{4} + \frac{y^2}{25}$ that is closest to the point $(3, 0, 0)$. 
So this seems like a pretty standard question of Lagrange multiplier, except I ran into some problems but cant figure out where I went wrong.
Attempt:
I maximuse $D^2$ instead of D to remove the square-root
1.) Constraint equation : $\frac{x^2}{4} + \frac{y^2}{25} - z$
2.) $D^2 = (x-3)^2 + y^2 + z^2$
3.) Do the usual Lagrange procedure
4.) $2z = -\lambda \\ 25y = \lambda y \\4(x-3) = \lambda x$
5.) Assume $x \neq 0, y\neq0$, we get $z = -\frac{25}{2}$
6.) This is where I got stuck, how can z be negative where is the sum of 2 positive number?
Which part of my attempt did I commit a mistake, and how do I rectify it?
Any insights and help is deeply appreciated.      
 [/INST] if $y=0$ then $z = \frac{x^2}{4}$ so $\lambda = -\frac{x^2}{2}$
so $$  4(x-3) =  -\frac{x^3}{2}  \\ \implies x^3+8x-24 = 0 \\ \implies(x-2)(x^2+2x+12)=0$$
so your solution is the point $(2,0,1)$
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