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Wednesday, July 1, 2015

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


Find the maximum of f(x)=x4x2x6+2x31 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}.


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