How to find what are the points closest to and farthest from (0,0) of ellipse $9x^2+4y^2=36$ using optimization? Please do not use Lagrange multipliers. Assume these have not been introduced and optimize. 
Edit: I try optimizing the squared distance formula using the equation as a constraint, but I only get one critical X out, so I don't know if that is the max or min, or how to find the other, since it's asking for both. 
Edit 2: This is a serious question. Please do not discriminate because I cannot get the editor to work. 
I get $x=-3/2$ and two corresponding y values, but is this the max or the min? Where do I plug this in to find out?
 A: This problem can be done without any calculations whatsoever just by understanding the graph of an ellipse in standard form. However just for fun I provide a calculation using calculus without the use of any transcendental functions.

We are trying to find the points on the ellipse, $ \frac{x^2}{4}+\frac{y^2}{9} = 1$, which are closest to and  farthest away from the origin. This means that we must optimize  the expression $d^2 = x^2+y^2$ subject to the given constraint. To do this we first introduce the parameterization,
$$ x = 2 \cdot \frac{2t}{1+t^2} \qquad y=3 \cdot \frac{1-t^2}{1+t^2}.$$
Substituting our parameterization into the expression for $d^2$ we get,
$$ d^2 = \frac{16 t^2 + 9(1-t^2)^2}{(1+t^2)^2}.$$  
Now we look for critical points by examining $(d^2)^\prime$,
$$ (d^2)' =  \frac{d}{dt} \frac{16 t^2 + 9(1-t^2)^2}{(1+t^2)^2} = \frac{40(t)(t^2-1)}{(t^2+1)^3},$$ 
the above expression is equal to zero when $t=0,1,-1$ or $t=\pm \infty$. To determine which values correspond to the minimum and maximum respectively we simply evaluate $x$ and $y$ for all the different values of $t$. We will manually pick out the values which are closest to and furthest from the origin.
$$x(0) = 0 \quad y(0)=3$$
$$x(1) = 2 \quad y(1)=0$$
$$x(-1) = -2 \quad y(-1)=0$$
$$x(\infty) = 0 \quad y(\infty) = -3$$
$$x(-\infty) = 0 \quad y(-\infty) = -3$$
The points furthest from the origin are $(0,3)$ and $(0,-3)$. The points closest to the origin are $(2,0)$ and $(-2,0)$.

Edit In response to the comment:
$t$ is just a parameter which labels the points on the ellipse. To understand the validity of this parametrization note that,
$$ \frac{1}{4}   \left( 2 \cdot \frac{2t}{1+t^2} \right)^2 
+  \frac{1}{9} \left( 3 \cdot \frac{1-t^2}{1+t^2}\right)^2
=1.$$
For a proof just google "rational points on the circle".
Its not necessary to do it this way I just happen to enjoy this rational parametrization. What you are suggesting about substituting $y^2$ as a function of $x$ into the $d^2$ equation will work just as well. Also you could parameterize with $x=2\sin \theta$ and $y=3\cos \theta$.
A: Note that the distance from the centre, which you want to measure is given by $r^2=x^2+y^2$ with $r\ge 0$ and maximising/minimising $r$ is therefore equivalent to maximising/minimising $r^2$.
To maximise $r^2$ note that $5x^2+4r^2=36$ so that $4r^2\le 36$ with equality, and a therefore a maximum, only if $x=0$.
To minimise $r^2$ note that $9r^2-5y^2=36$ so that $9r^2\ge 36$ with equality, and a minimum, only if $y=0$.
