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Linking Road

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A road is to be build, linking a village ALT= to a town ALT=. With distance ALT= from the village, a motorway already goes to the town. The point ALT= on the motorway that is closest to the village, lies ALT= off the town.

The straight road is to be build in such a way that the town can be reached as quickly as possible. Hereby one can assume an average speed ALT=km/h on the motorway as well as an average speed ALT=km/h on the new road.

ALT=

Determine the distance ALT= between ALT= and the point where the new road is to meet the motorway by minimizing the time

$\displaystyle t(x)=\sqrt{a^2+x^2}/v_n + (b-x)/v_a $

required to drive from the village to the town.

$ t^\prime(x)=0$

$\displaystyle t^\prime(x) = \frac{x}{\sqrt{a^2+x^2}v_n}-\frac{1}{v_a} \overset{!}{=} 0\,. $

Thus

$\displaystyle v_a x = v_n \sqrt{a^2+x^2} \quad \Leftrightarrow \quad x_m = a/\sqrt{3} $

is the local extremum in $ [0,\,b]$. Whether this is a minimum can be controlled by comparing the driving time with that for the interval's boundary points:

$\displaystyle t(a/\sqrt{3}) = \frac{\sqrt{3}a+b}{v_a}\,, \qquad t(0) = \frac{2a+b}{v_a}\,, \qquad t(b)=\frac{2\sqrt{a^2+b^2}}{v_a} $

Obviously it is $ t(0) \geq t(x_m)$. For $ x=b$ it is

$\displaystyle t(b) \geq t(x_m)$ $\displaystyle \Leftrightarrow$ $\displaystyle 2\sqrt{a^2+b^2} \geq \sqrt{3}a+b$  
  $\displaystyle \Leftrightarrow$ $\displaystyle 4a^2+4b^2 \geq 3a^2+2\sqrt{3}ab+b^2$  
  $\displaystyle \Leftrightarrow$ $\displaystyle a^2-2\sqrt{3}ab+3b^2 =(a-\sqrt{3}b)^2 \geq 0\,.$  

The latter unequality is always true, thus $ x_m$ is optimal if $ b \geq a/\sqrt{3}$. Otherwise $ x_m$ is not in $ [0,b]$, then the boundary point ALT= is the minimum, which would be a direct connection to the town.

(Authors: Höllig/Hörner/Abele)
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  automatisch erstellt am 8.  4. 2008