## Wednesday, July 1, 2015

### Optimization Contest Problem: Find the maximum of f(x) (Heuristic Solution)

Find the maximum of $f(x) = \dfrac{x^4-x^2}{x^6+2x^3-1}$ where $x>1$.

If we want to maximize $f(x)=\dfrac{x^4-x^2}{x^6+2x^3-1}=\dfrac{1}{\left(\dfrac{x^6+2x^3-1}{x^4-x^2}\right)}$, this could be done if we are to find the minimum value for the expression $\dfrac{x^6+2x^3-1}{x^4-x^2}$.

Note that

\begin{align*}\dfrac{x^6+2x^3-1}{x^4-x^2}&=x^2+1+\dfrac{2x^3+x^2-1}{x^4-x^2}\\&=x^2+1+\dfrac{2x^3}{x^2(x^2-1)}+\dfrac{x^2-1}{x^2(x^2-1)}\\&=x^2+1+\dfrac{2x}{x^2-1}+\dfrac{1}{x^2}\\&=\left(x-\dfrac{1}{x}\right)^2+\dfrac{2}{\left(x-\dfrac{1}{x}\right)}+3\\&=\left(x-\dfrac{1}{x}\right)^2+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}+3---(*)\end{align*}

And since the term $x-\dfrac{1}{x}$ is a positive in the given domain, we then are allowed to apply the AM-GM to the first three terms of the expression (*) and that gives

$\left(x-\dfrac{1}{x}\right)^2+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}\ge 3$

Equality occurs when $\left(x-\dfrac{1}{x}\right)^2=\dfrac{1}{\left(x-\dfrac{1}{x}\right)}$, i.e.

$\left(x-\dfrac{1}{x}\right)^3=1$

$x-\dfrac{1}{x}=1$

$x^2-x-1=0$

$\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0$

$x=\dfrac{1+\sqrt{5}}{2}$

Therefore, the minimum value for $\dfrac{x^6+2x^3-1}{x^4-x^2}$ is $3+3=6$ and that results in a maximum of

$f(x)=\dfrac{x^4-x^2}{x^6+2x^3-1}=\dfrac{1}{\left(\dfrac{x^6+2x^3-1}{x^4-x^2}\right)}$ as $\dfrac{1}{6}$.