Given the points $A(3,3)$, $B(0,1)$ and $C(x,0)$ where $0 < x < 3$, $AC$ is the distance between $A$ and $C$ and $BC$ is the distance between $B$ and $C$. What is x for the distance $AC + BC$ to be minimal?

What have I done? I defined the function $AC + BC$ as:

$\mathrm{f}\left( x\right) =\sqrt{{1}^{2}+{x}^{2}}+\sqrt{{3}^{2}+{\left( 3-x\right) }^{2}}$

And the first derivative:

$\mathrm{f'}\left( x\right) =\frac{x}{\sqrt{{x}^{2}+1}}+\frac{3-x}{\sqrt{{x}^{2}-6\,x+18}}$

We need to find the values for $\mathrm{f'}\left( x\right) = 0$, so by summing and multiplying both sides I got to the equation:

$2x^4 - 12x^3 + 19x^2 -6x+18 = 0$

But I don't think the purpose should be to solve a 4th grade equation, there should be another way I'm missing..

  • $\begingroup$ Looks like a Lagrange multiplier problem. $\endgroup$ – math-visitor Jun 3 '12 at 10:51

Norbert and Ross Millikan have already suggested one slick solution, and Gerry Myerson the even slicker solution, but you can do it purely algebraically: you ran into trouble because your derivative isn’t quite right. It should be


with the second term negative from the chain rule applied to $3-x$. Setting that to $0$ and doing a little algebra, we get

$$\begin{align*} x\sqrt{18-6x+x^2}&=(3-x)\sqrt{1+x^2}\;,\\ 18x^2-6x^3+x^4&=(3-x)^2(1+x^2)\;,\\ 18x^2-6x^3+x^4&=9-6x+10x^2-6x^3+x^4\;,\\ 18x^2&=9-6x+10x^2\;,\\ 8x^2+6x-9&=0\;,\text{ and}\\ (4x-3)(2x+3)&=0\;. \end{align*}$$

The desired solution is clearly $x=\dfrac34$.


Use physical analogy. Consider $ACB$ as the path of light travelling from $A$ to $B$ with reflection at point $C$. As we all know light travels along the shortest path, and angle of incidence equals angle of reflection. So you had to choose point $C$ such that angle between $x$-axis and $AC$ must be equal to angle between $x$-axis and $CB$

enter image description here

Since angles $ACO$ and $BCD$ are equal we can say that their tangents are equal. Hence $$ \frac{1}{x}=\frac{3}{3-x} $$ and we get $x=3/4$

  • $\begingroup$ Thanks, +1 but I'll accept Brian's answer since it's what my assignment was about $\endgroup$ – Schiavini Jun 3 '12 at 11:05
  • $\begingroup$ Not at all.${}{}{}$ $\endgroup$ – Norbert Jun 3 '12 at 11:07
  • $\begingroup$ @Norbert: Great. nothing can describe the problem but a graphical view. $\endgroup$ – mrs Jun 3 '12 at 11:09
  • $\begingroup$ @Norbert, I'm learning derivatives $\endgroup$ – Schiavini Jun 3 '12 at 14:18

Let $B'=(0,-1)$. Do you see that $AC+BC=AC+B'C$? Do you see where to put $C$ to minimize $AC+B'C$?

  • $\begingroup$ This is a great answer, +1 but I'll accept Brian's answer since it's what my assignment was about $\endgroup$ – Schiavini Jun 3 '12 at 11:05

The following little trick (method?) is useful in many places. Let $y=3-x$. We want to minimize $\sqrt{1+x^2}+\sqrt{3^2+y^2}$, where $x+y=3$.

Differentiate with respect to $x$. Using the fact that $\frac{dy}{dx}=-1$, we arrive at the equation $$\frac{x}{\sqrt{1+x^2}}=\frac{y}{\sqrt{9+y^2}}.$$ Note that this is (apart from a little minus sign problem) the equation you arrived at, with $3-x$ replaced by $y$. Square both sides, cross-multiply. We get $$x^2(9+y^2)=y^2(1+x^2),$$ which simplifies to $9x^2=y^2$.

No fourth degree equation here! Since $x$ and $y$ are non-negative, we get $y=3x$, a linear equation. Now from $x+y=3$ we obtain $x=3/4$.


Hint: you might look at Wikipedia on angle of reflection


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