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?
Best Answer
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 $y=3\cos \theta$.