Find minimal value of $f(x) = \sqrt{5x^2 - 40x + 85} + \sqrt{5x^2 - 24x + 53}$.

I can solve it using derivatives. Is there any other way to solve it? For example using some popular inequalities?



We can write $$\frac{f(x)}{\sqrt 5}=\sqrt{(x-4)^2+(0-1)^2}+\sqrt{\left(x-\frac{12}{5}\right)^2+\left(0-\frac{11}{5}\right)^2}.$$

This represents the sum of the distance between $(x,0)$ and $(4,1)$ and the distance between $(x,0)$ and $\left(\frac{12}{5},\frac{11}{5}\right)$.

  • $\begingroup$ So, $\sqrt{(x-4)^2+(0-1)^2}+\sqrt{\left(x-\frac{12}{5}\right)^2+\left(0-\frac{11}{5}\right)^2}$ from inequality of triangle is higher than 2. What's next? $\endgroup$ – user128409235 Nov 5 '15 at 19:04
  • 3
    $\begingroup$ @user128409235: The sum is minimum when $(x,0)$ is on the line passing through $(4,1)$ and $(12/5,\color{red}{-}11/5)$. (Why?) $\endgroup$ – mathlove Nov 5 '15 at 19:08
  • $\begingroup$ I understand. Thank you :). $\endgroup$ – user128409235 Nov 5 '15 at 19:19
  • $\begingroup$ @user128409235: You are welcome. $\endgroup$ – mathlove Nov 5 '15 at 19:20

As suggested by mathlove, if we set $A=\left(\frac{12}{5},\frac{11}{5}\right), B=(x,0)$ and $C=\left(4,1\right)$, we have: $$ f(x) = \sqrt{5}\left( AB+BC \right) $$ that is minimized when $B$ belongs to the $AC$-line.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.