Optimization problem on function

[juantheron]

Maximum value of expression $\displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1}\;,$ where $x>1$

 

[anemone]

My solution:

Spoiler

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

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

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

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 $\dfrac{1}{6}$

.

 

[anemone]

Hi jacks again, :)

I want to apologize for not formulating my solution above in a very good way.:(

I forgot to mention that the term $\dfrac{1}{\left(x-\dfrac{1}{x}\right)}$ is a positive in the given domain where $x>1$, therefore, I could then apply the AM-GM on those terms to look for the minimum value for $\left(x-\dfrac{1}{x}\right)^2+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}$.

Sorry about that.

 

[juantheron]

Thanks anemone for Nice Solution::

My Solution is Similar To Yours::

Spoiler

Given \(\displaystyle \displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1}\;,\) Where \(\displaystyle x>1\)

So We can Simplify \(\displaystyle \displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1} = \frac{x^3\cdot \left(x-\frac{1}{x}\right)}{x^3\cdot \left(x^3-\frac{1}{x^3}\right)+2} = \frac{\left(x-\frac{1}{x}\right)}{\left(x^3-\frac{1}{x^3}\right)+2}\)

Now Let \(\displaystyle u = \left(x-\frac{1}{x}\right)>0\;,\left(x^3-\frac{1}{x^3}\right) = \left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\;,x>1.\)

So \(\displaystyle \displaystyle f(u) = \frac{u}{u^3+3u+2}=\frac{u}{u^3+u+u+u+1+1}\leq \frac{u}{6\sqrt[6]{u^3\cdot u \cdot \cdot u \cdot u\cdot 1\cdot 1}} = \frac{1}{6}\)

Using \(\displaystyle \bf{A.M\geq G.M}\) and above equality hold when \(\displaystyle \displaystyle u = 1\Rightarrow \left(x-\frac{1}{x}\right) = 1\Rightarrow x= \frac{\sqrt{5}+1}{2}\)

 

[CellCoree]

I would approach this optimization problem by first understanding the function and its properties. The given expression is a rational function, which means it is a ratio of two polynomials. It is also specified that the value of x must be greater than 1.

To find the maximum value of the function, we can use calculus and take the derivative of the function with respect to x. This will give us the critical points of the function where the maximum or minimum values can occur.

Taking the derivative of the given function, we get:

$\displaystyle \frac{d}{dx}f(x) = \frac{(x^4-x^2)(6x^5+6x^2)-(x^6+2x^3-1)(4x^3-2x)}{(x^6+2x^3-1)^2}$

Setting this derivative equal to 0 and solving for x, we get the critical point at x = 1. This means that the maximum value of the function occurs at x = 1.

To confirm this, we can also take the second derivative of the function and evaluate it at x = 1. If the second derivative is positive, then the critical point at x = 1 is a minimum and if it is negative, then it is a maximum.

$\displaystyle \frac{d^2}{dx^2}f(x) = \frac{(x^6+2x^3-1)^2(12x^4+12x)-(x^4-x^2)(2(6x^5+6x^2)+4x^3-2x)(2(6x^3+2x)-4)}{(x^6+2x^3-1)^4}$

Evaluating this at x = 1, we get a positive value. This confirms that x = 1 is a local maximum point.

Therefore, the maximum value of the given expression is $\displaystyle f(1) = \frac{1^4-1^2}{1^6+2(1^3)-1} = \frac{0}{2} = 0$.

In conclusion, the optimization problem on the given function has a maximum value of 0, which occurs at x = 1. This solution can also be verified by plotting the function and observing that it has a local maximum at x = 1.