A road between two cities

linewirefx · March 13th 2008, 11:27 PM

I appreciate any help that I get, I just can't figure out how to do this problem. If anyone else knows, I would love some input!

Two cities lie south of a straight road but they are not connected to the road or to each other. The people of the city decided to build two roads, one from each city to the existing road. We do not have exact values to work with, so you need to assign letters to the distances you need for the problems. Distance from one city to the road; distance from other city to the road; Distance on the road they are from each other. The answer will be in the terms of these three unknown constants.

Use Calculus - Optimization - to find the shortest route connecting these two cities via the existing road.

Here is a picture of the problem >>
http://www.pplunder.com/hdbin/public/8637city.JPG

**galactus** · March 14th 2008, 04:19 PM

I f I am interpreting correctly, you must find a point P which minimizes the distances from the cities to the main road.

You can do this with similar triangles.

$\frac{x}{a}=\frac{c-x}{b}$

Solve for x and we get $x=\frac{ac}{a+b}$

You could derive a formula using Pythagoras and then differentiate, set to 0 and solve for x, but you'll get the same solution only with a lot more work.

$D=\sqrt{a^{2}+x^{2}}+\sqrt{b^{2}+(c-x)^{2}}$

$D'=\frac{x\sqrt{x^{2}-2cx+b^{2}+c^{2}}+\sqrt{x^{2}+a^{2}}(x-c)}{\sqrt{(x^{2}+a^{2})(x^{2}-2cx+b^{2}+c^{2})}}$

$x\sqrt{x^{2}-2cx+b^{2}+c^{2}}+\sqrt{x^{2}+a^{2}}(x-c)=0$

$x^{2}(x^{2}-2cx+b^{2}+c^{2})=(x^{2}+a^{2})(x-c)^{2}$

$(b^{2}-a^{2})x^{2}+2a^{2}cx-a^{2}c^{2}=0$

Factor:

$((a-b)x-ac)(ac-(a+b)x)$

$(a-b)x-ac=0; \;\ x=\frac{ac}{a-b}$

$ac-(a+b)x=0; \;\ \boxed{x=\frac{ac}{a+b}}$

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