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Let $a,b,c \in \mathbb{R}^+$. How can we find the minima of the following function: $$ f(x) = \sqrt{a^2+x^2} + \sqrt{(b-x)^2 + c^2} $$ without calculating the derivative?

I showed that $a+c$ is a minorant, it is à minima iff $b=0$. But I couldn't find the minima for $b\neq 0$.

odwimi
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2 Answers2

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Note tha $\sqrt{a^2+x^2}$ is the distance from $p_1=(0,a)$ to $(x,0)$ and that $\sqrt{(b-x)^2 + c^2}$ is the distance from $p_2=(b,-c)$ to $(x,0)$. So all you meed to find is where the line crossing $p_1$ and $p_2$ is crossing the $x$ axis. This will be the point where your function achieves its minima.

Peter Müller
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Here is a solution based on the triangle inequality for the Euklidean distance in the plane $\mathbb{R}^2$:

  • Euklidean distance: $u=\binom{u_1}{u_2}\Rightarrow |u| = \sqrt{u_1^2 + u_2^2}$
  • Triangle inequality: $u,v \in \mathbb{R}^2 \Rightarrow |u+v| \leq |u| + |v|$ with equality iff $u = tv$ or $v=tu$ for some $t\geq 0$.

Now, set $u = \binom{x}{a}, v= \binom{b-x}{c}$, and you get

\begin{eqnarray*} \sqrt{x^2+a^2} + \sqrt{(b-x)^2+c^2} & = & |u| + |v| \\ & \geq & |u+v| \\ & = & \sqrt{b^2 + (a+c)^2} \end{eqnarray*}

For example, for the case $a,b,c>0$ you get equality iff

$$t\binom{x}{a} = \binom{b-x}{c} \Leftrightarrow t= \frac ca, x=\frac{ab}{a+c}$$

I leave the other cases up to you.

trancelocation
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