4
$\begingroup$

Let $E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}.$ Find the $x$ value for when this expression take it's minimum value.

Using the derivative it's pretty hard... and I think not indicated for this problem... I thought about using the inequality of means. First, rewrite:

$E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}$ as $E(x)=\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}$..

But I don't know how to continue...

$\endgroup$
10
$\begingroup$

Note that $$ \sqrt{\left(x-1\right)^2+4}=\sqrt{\left(x-1\right)^2+\left(0-\left(-2\right)\right)^2} $$ is exactly the distance between point $A:\left(x,0\right)$ and point $B:\left(1,-2\right)$, and that $$ \sqrt{\left(x-4\right)^2+9}=\sqrt{\left(x-4\right)^2+\left(0-3\right)^2} $$ is exactly the distance between point $A:\left(x,0\right)$ and point $C:\left(4,3\right)$. Therefore, you are to find some point $A$ on the $x$-axis such that $$ d(A,B)+d(A,C) $$ is minimized (i.e., the total distance of the red segments in the figure below is minimized), where $d(A,B)$ denotes the distance between points $A$ and $B$. enter image description here

Gratefully, it is obvious that, as $A$ falls on the dashed blue segment, the total distance witnesses its minimum, since this dashed blue segment yields the minimum distance between points $B$ and $C$. Due to the triangular inequality, any other position of $A$ would make $$ d(A,B)+d(A,C)>d(B,C). $$ Therefore, the very $A$ that minimizes the total distance should be the point on the dashed blue segment. Then combined with the fact that $A$ is a point on the $x$-axis, this $A$ would have no choice but to be $\left(11/5,0\right)$.

$\endgroup$
  • $\begingroup$ So, $\frac {11 }5$? $\endgroup$ – C. Cristi Apr 26 '18 at 11:36
  • $\begingroup$ @C.Cristi: Perfect! $x=11/5$ seems to be the answer. $\endgroup$ – hypernova Apr 26 '18 at 11:39
  • $\begingroup$ Also can you be more explicit with why you did what you did? And what made you think of that? and do you think it would work using the derivative? Also Why the intersection of the line that crosses $(1,-2), (4,3)$ gives me my minimum of that expression? $\endgroup$ – C. Cristi Apr 26 '18 at 11:39
  • $\begingroup$ @C.Cristi: Er... I thought you were seeking for some shortcut. Hopefully this type of questions are designed to have a default shortcut, i.e., considering the geometric interpretation of $E(x)$. So as you realize that you need to minimize the total distance between the three points, you are approaching the answer. $\endgroup$ – hypernova Apr 26 '18 at 11:44
  • 1
    $\begingroup$ @C.Cristi: Exactly. But there are only a few interpretations in total, and you will not miss it. This method applies as long as the cost function is of the form $\sqrt{a_1x^2+b_1x+c_1}+\sqrt{a_2x^2+b_2x+c_2}$. $\endgroup$ – hypernova May 20 '18 at 10:24
1
$\begingroup$

Using Minkowski's inequality: $$\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}=\\ \sqrt{(x-1)^2+2^2}+\sqrt{(4-x)^2+3^2}\ge \\ \sqrt{(x-1+4-x)^2+(2+3)^2}=\sqrt{3^2+5^2}=\sqrt{34}.$$ The equality occurs when $(x-1,2)||(4-x,3)$, that is: $$\frac{3}{2}(x-1)=4-x \Rightarrow x=\frac{11}{5}.$$

$\endgroup$

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.