Maximizing a Complex Function: Solving the Optimization Challenge II

[anemone]

Find the maximum value of the function $\sqrt{x^4-9x^2-12x+61}-\sqrt{x^4-x^2+1}$.

 

[juantheron]

My Solution:

Spoiler

Given $$\displaystyle \sqrt{x^4-9x^2-12x+61}-\sqrt{x^4-x^2+1} = \sqrt{\left(x^2-5\right)^2+\left(x-6\right)^2}-\sqrt{\left(x^2-1\right)^2+(x-0)^2}$$

Now Using Triangle Inequality::

$$\displaystyle \sqrt{\left(x^2-5\right)^2+\left(x-6\right)^2}-\sqrt{\left(x^2-1\right)^2+(x-0)^2}\leq \sqrt{\left(x^2-5-x^2+1\right)^2+\left(x-6-x\right)^2} = 2\sqrt{13}$$

and equality hold when $$\displaystyle \frac{x^2-5}{x-6} = \frac{x^2-1}{x}$$

 

[anemone]

My Solution:

Spoiler

Given $$\displaystyle \sqrt{x^4-9x^2-12x+61}-\sqrt{x^4-x^2+1} = \sqrt{\left(x^2-5\right)^2+\left(x-6\right)^2}-\sqrt{\left(x^2-1\right)^2+(x-0)^2}$$

Now Using Triangle Inequality::

$$\displaystyle \sqrt{\left(x^2-5\right)^2+\left(x-6\right)^2}-\sqrt{\left(x^2-1\right)^2+(x-0)^2}\leq \sqrt{\left(x^2-5-x^2+1\right)^2+\left(x-6-x\right)^2} = 2\sqrt{13}$$

and equality hold when $$\displaystyle \frac{x^2-5}{x-6} = \frac{x^2-1}{x}$$

Well done, jacks!(Yes) That is the trick to make this challenge as simplest as possible, and thanks for participating!:)