Minimizing $\sqrt{(x+3)^2 + 49} + \sqrt{(x-5)^2 + 64}$

Question

What is the minimum value of $$\sqrt{(x+3)^2 + 49} + \sqrt{(x-5)^2 + 64}$$

I tried getting the first derivative, but I can't solve the equation when I put $y = 0$.

Methods without using calculus are also welcome.

Answer 1

Here's a solution without words ...

Answer 2

Using Triangle Inequality

$$\sqrt{(x+3)^2+7^2}+\sqrt{(x-4)^2+8^2}\geq \sqrt{(3+x+x-7)^2+(7+9)^2}$$

Answer 3

By Minkowski (triangle inequality) we obtain: $$\sqrt{(x+3)^2+7^2}+\sqrt{(x-4)^2+8^2}\geq \sqrt{(x+3+4-x)^2+(7+8)^2}=\sqrt{274}$$ The equality occurs for $(x+3,7)||(4-x,8),$ which says that we got a minimal value.

Answer 4

$$f'(x)=\frac{x+3}{\sqrt{(x+3)^2+49}}+\frac{x-5}{\sqrt{(x-5)^2+64}}$$ $$f'(x)=0\quad\Rightarrow$$ $$15x^2+874x-649=0$$ $$x=\frac{11}{15},\quad x=-59$$ Then $$f_{min}=f\left(\frac{11}{15}\right)=17$$

Answer 5

This function $$y=\sqrt{(x+3)^2 + 49} + \sqrt{(x-5)^2 + 64}$$ is symmetric about $x=\dfrac{11}{15}$, then since that is sum of two hyperbola, it's shape is like a symmetrical parabola so the minimum accurs in $x=\dfrac{11}{15}$, it is $17$.