Olympiad Math Problem: Find the maximum and minimum of P (First Attempt)

Find the minimum and maximum of $P=\dfrac{y−x}{x+8y}$ for all real $x$ and $y$ that satisfy the equation $y^2(6-x^2)-xy-1=0$.

On one hand, this is not a unmanageable Olympiad Mathematics optimization problem, on the other hand, this problem allows us to show students how powerful algebraic manipulation is when we use it diligently and how effective the accurate solution we could have arrived compared to all the alternatives.

One might want to try the typical old method (calculus approach) to tackle it, let's see if such an approach would be fruitful.

First off, we need to rewrite the function of $P$ such that it is in terms of only one variable for easy referencing and easy working.

From the given equation where both real $x$ and $y$ that satisfy, it would be hard to rewrite either $x$ in terms of $y$ or the other way round.

$y^2(6-x^2)-xy-1=0$

$\begin{align*}y&=\dfrac{x\pm \sqrt{x^2-4(6-x^2)(-1)}}{2(6-x^2)}\\&=\dfrac{x\pm \sqrt{24-3x^2}}{2(6-x^2)}\end{align*}$

What do you think now? Do you think you want to proceed to find the expression for $y$? Let's assume you want to go ahead to do that...you would end up with the following:

Now comes the time to make up your mind and make the good decision based on your finding.

Would you think it is useful to substitute the above expression for $y$ into the target expression of $P$?

$\begin{align*}P&=\dfrac{y−x}{x+8y}\\&=\dfrac{\dfrac{x\pm \sqrt{24-3x^2}}{2(6-x^2)}−x}{x+8\left(\dfrac{x\pm \sqrt{24-3x^2}}{2(6-x^2)}\right)}\\&=\dfrac{x\pm \sqrt{24-3x^2}-x(2)(6-x^2)}{2x(6-x^2)+8x\pm 8\sqrt{24-3x^2}}\\&=\dfrac{2x^3-11x\pm \sqrt{24-3x^2}}{20x-2x^3\pm \sqrt{24-3x^2}}\end{align*}$

This looks horrible and if you are persistent, you might want to try it so to differentiate $P$ w.r.t. $x$, we get:

$\small\begin{align*}P'&=\dfrac{(20x-2x^3\pm \sqrt{24-3x^2})\left(6x^2-11\pm\dfrac{(-6x)}{\sqrt{24-3x^2}}\right)-(2x^3-11x\pm \sqrt{24-3x^2})\left(20-6x^2\pm\dfrac{(-6x)}{\sqrt{24-3x^2}}\right)}{(20x-2x^3\pm \sqrt{24-3x^2})^2}\end{align*}$

What about it now? What's your decision? Do you still want to continue with what you have left off?

It's a choice you make, but if I were you, I could stop and try to approach the problem differently. The first derivative is necessarily hard to work with and it seems it's virtually impossible to determine the critical points from it.

I will post in the next blog post the different way of approaching the problem and I hope you would attempt at a solution and remember, you learn best from your mistake.