find a point on ellipse closest to origin Find the points on the ellipse $2x^2 + 4xy + 5y^2 = 30$ closest and farthest from origin. How to do this problem? I know how to find a closest point if $z = f(x,y)$ is given, however, this is 2 dimensional.

If Lagrange multipliers are not yet part of the toolkit, or even if they are, we can use a perpendicularity argument. 
Visualize circles $x^2+y^2=r^2$, where $r$ is a variable that starts very close to $0$, and increases.
There is a *smallest*value of $r$ such that $x^2+y^2=r^2$ meets our ellipse, and a biggest. These are what we are looking for.  At these values of $r$, the circle $x^2+y^2=r^2$ is tangent to our ellipse. 
Let the point of tangency be $P=(a,b)$. Then
$$2a^2+4ab+5b^2=30.\tag{$1$}$$
The line $OP$ is perpendicular to the tangent at $(a,b)$ of the circle, and hence of the ellipse.
Now we can use either one variable calculus (implicit differentiation) or partial derivatives. If $f(x,y)=2x^2+4xy+5y^2-30$, we have $f_x(x,y)=4x+4y$ and $f_y(x,y)=4x+10y$. 
The perpendicularity condition is that the dot product of $(4a+4b, 4a+10b)$ and $(a,b)$ is $0$. So we get
$$a(4a+4b)+b(4a+10b)=4a^2+14ab+4b^2=0.\tag{$2$}$$
Solve, using the Qadratic Formula. We get
$$a=\left(\frac{-7\pm\sqrt{33}}{4}\right)b.$$
Now substitution in Equation $(1)$gives our candidates.
