Question: Points P and Q are chosen on the curve X^4 + 16Y^4 = 1 in
such a way that the distance PQ is as large as possible. Find that
distance.

Key points: i am in first year differential calculus, i do not have
any knowledge past simple differentiation, Newton Rhapson method and
linear approximation. These are the only tools available to me (no
Lagrangage numbers or anything).

The answer to the problem is 17^(1/4)...does anyone have a simple
detailed solution for someone who is a beginner at calculus(!)? Ive
tried this for about 6 hours.

Suggestion: Maximize X^2 + Y^2, subject to the constraint given by the
equation of your curve.  In other words, eliminate one of the
variables X or Y (probably X would be the easiest to eliminate). 
[Note:  As a further simplification, your expression to maximize will
actually be a function of Y^2 only, so you might avoid a little work
by substituting Z = Y^2 at this point.]

This gives you the point(s) on the curve farthest from the origin, and
so the pair of _symmetric_ points P,Q farthest apart.  Now argue that
for any two points _not_ symmetrically placed on the curve, a greater
distance between them can be obtained by moving one of them to a
position opposite the one furthest from the origin.  Thus the pair of
points P,Q furthest apart overall must be symmetric with respect to
the origin.

regards, mathtalk-ga