Go the distance

Consider the figure below. Title of picture $AB=\sqrt{(x+6)^2+25}$ and $BC=\sqrt{(x-6)^2+121}$ . The sum $AB+BC$ is minimized when $x$ is such that $AB+BC=AC=20$ .

$Let\quad us\quad define\quad the\quad complex\quad numbers\quad ,\\ z1\quad =\quad 11\quad -\quad (x-6)i\quad \\ and\quad z2\quad =\quad 5\quad +\quad (x+6)i\\ \\ Therefore,\quad \left| z1 \right| =\sqrt { 11^{ 2 }\quad +\quad (x-6)^{ 2 }\quad } \quad \& \quad \left| z2 \right| =\sqrt { 5^{ 2 }\quad +\quad (x+6)^{ 2 } } \\ \\ Now\quad applying\quad the\quad triangle\quad inequality\quad of\quad complex\quad numbers,\quad \\ \left| z1 \right| \quad +\quad \left| z2 \right| \quad \ge \quad \left| z1\quad +\quad z2 \right| \\ \\ \Rightarrow \quad \left| z1\quad +\quad z2 \right| \quad =\quad \sqrt { { 16 }^{ 2 }\quad +\quad { 12 }^{ 2 } } \quad =\quad 20.\\ Which\quad is\quad the\quad minimum\quad value.$

$\sqrt{(x + 6)^2 + 25}$ is the distance between the points $(x , 3)$ and $(-6, 8)$

and

$\sqrt{(x - 6)^2 + 121}$ is the distance between the points $(x , 3)$ and $(6, -8)$

by Distance Formula

(This is because $(3 - 8)^2 = 25$ and $(3 + 8)^2 = 121$ )

Therefore, the minimum distance will be achieved if and only if all of the three points lie on the same line.

Therefore, the minimum distance is the distance between the points $(-6, 8)$ and $(6, -8)$ which is $\boxed{20}$

I found a way but some may find it too blunt, but i rather pick this standard way when i'm stuck. (For people who couldn't approach using the solution by Dan Lawson) ;)

Just think of finding the minimum of a normal function $f(x)$ , we just find where $f'(x)$ is 0 and there you get maximum or minimum. Let $f(x)=\sqrt{(x+6)^{2}+25} +\sqrt{(x-6)^{2}+121} \frac{}{}$ So derivative of $f(x)$ is $f'(x)=\frac{x+6}{\sqrt{(x+6)^{2}+25}}+\frac{x-6}{\sqrt{(x-6)^{2}+121}}=0$ Its quite easy to solve after squaring on both sides and solving the quadratic equation $4x^{2}+73x+144=0$ Roots $\frac{-9}{4}$ and -16,
Value of the $f(x)$ at the roots = $20,5\sqrt{3}$

So Minimum value is $\boxed{20}$

First think graphically. It is to find a point on x-axis such that its distances from P(-6,5) and Q(6,11) add up to a minimum. And then, geometry comes in. If you prefer, it is the optics of physics. The minimum distance is the distance between P and the mirror-image of Q in x-axis, i.e.(6,-11). Now (-6-6)^2 + (5-(-11))^2 is 20^2. Hence the minimum value is 20.

The minimum sought is the reflection of Point P into M. See the image therefore by similar triangles we get. (6+x)/11=(6-x)/5 resolving for x we get x=2.25 and the minimum equals 2 0

We can use calculus....though it will be a little bit lengthy...The 1st derivative of the function is 4x^2+73x+144....We set it to be 0 and solving for x we find x=-9/4, 16....The 2nd derivative of the function is 8x+73...plugging in -9/4 here gives lesser value than 16...so plugging -9/4 in the main function gives the minimum value and that is 20...

By the Minkowski's Inequality, $\sqrt {(x+6)^2 + 5^2} + \sqrt {(x-6)^2 + 11^2} \geq \sqrt {((x+6)-(x-6))^2 + (5+11)^2} = 20$ .

I'll leave the equality case to be found out.

just the use of fermats principle!!

required-- distance between the points:: (-6,5) and (6,-11) for the sum to be min. which is =20

First, draw a plane coordinate system like this: The distance between $(-6,5)$ and $(x,0)$ is $\sqrt[2]{(x-(-6))^2+(5-0)^2} = \sqrt[2]{(x+6)^2+25}$ . The distance between $(x,0)$ and $(6,-11)$ is $\sqrt[2]{(x-6)^2+(0-(-11))^2} = \sqrt[2]{(x-6)^2+121}$ . The minimum value of the original formula is the distance between two blue dots: $\sqrt[2]{((-6)-6)^2+(5-(-11))^2} = \sqrt[2]{12^2+16^2} = \sqrt[2]{144+256} = \sqrt[2]{400} = 20$

https://www.geogebra.org/m/wTcucKp5

Using the diagram by Dan Lawson $\sqrt{AC}=\sqrt{12^2+16^2}=20.$

The best way to solve this problem is using MatLab to graph the whole function from -100 to 100.

a lot of people have brilliant solutions and i am here hacking the system graphed it on desmos and figured it out

The derivative is 0 at x=-2.25. The value at this point is 20 and a bigger value can easily be found to show that it is minima.