Finding the distance between a point and a parabola with different methods I'm trying to find the shortest distance from point $(3,0)$ to the parabola $y= x^2$ using the method of Lagrange Multipliers (my practice), and by "reducing to unstrained problem in one variable" (assignment). (I think the sheet might have meant unrestrained, but in either case, I don't know how to go about it). 
How would I go about this?
For Lagrange Multipliers, I think the set-up (from what I've read) is something like $(x-3, y) = λ(2x, ??)$ But I'm unsure where to get the y coordinate for the lambda side. I then solve for λ. From there on, I have no idea what to do.
I would greatly appreciate if someone could point me in the right direction or walk me through this. 
Thank you very much in advance.
 A: Draw a picture. 
For either method, we want to minimize $(x-3)^2+y^2$, subject to $y=x^2$. 
Substituting $x^2$ for $y$ in $(x-3)^2+y^2$, we find that we want to minimize $f(x)=(x-3)^2+x^4$. Note that $f'(x)=2(x-3)+4x^3$. The critical points of $f(x)$ are where $2(x-3)+4x^3=0$, or equivalently
$$2x^3+x-3=0.$$
In general cubic equations are unpleasant to solve. However, the above equation has the obvious root $x=1$. And since $f''(x)=3x^2+1\gt 0$, the function $f'(x)$ is increasing, and therefore can only have one zero. By the geometry, there is at least one point on the parabola at minimum distance from $(3,0)$, so we have found it. The minimum distance is the distance from $(3,0)$ to $(1,1)$, which is $\sqrt{5}$.
We now sketch the Lagrange multipliers approach. The Lagrangian is
$$(x-3)^2+y^2-\lambda(y-x^2).$$
Take the partial derivatives, and set them equal to $0$. We get $2(x-3)+2\lambda x=0$ and $2y-\lambda=0$. 
So $\lambda=2y$. Substituting in the first equation we get $2(x-3)+4yx=0$. Putting $y=x^2$ we obtain $2(x-3)+4x^3=0$, and we are at an equation we have already dealt with.
