# Finding the angle of a by-pass road given the lengths of two roads forming a right angle

The north-east section of a municipality is a rectangle whose 17 km west boundary is Highway 7 and whose 20 km south boundary is Highway 15. A by-pass road is planned to alleviate traffic congestion. If the by-pass is a straight road at what angle to Highway 15 should the new road be made to achieve the shortest route?

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What have you tried? –  Alexei Averchenko Dec 1 '12 at 16:35

A picture should accompany the solution. I will instead use a thousand words.

Let the "north-east section of a municipality" have corners $OPQR$, where $O$ is at the south-west corner, and the points $OPQR$ travel counterclockwise. So if as natural we call $O$ the origin, then $P$ is on the positive $X$-axis, and $R$ is on the positive $y$-axis.

We want to choose a point $X$ on the $x$-axis, somewhere east of town, and a point $Y$ on the $y$-axis, somewhere north of town, so that the line segment $XY$ is as short as possible, and does not meet town (it bypasses town).

To minimize the length of $XY$, it is clear that the line $XY$ must just "clip" town, that is, pass through the north-east corner $Q$.

Let us look at the consequences of making the angle $OXY$ equal to $\theta$. Then $$\frac{17}{XQ}=\sin \theta.$$ Now look at the part $YQ$ of the bypass $XY$. By the same reasoning we have $$\frac{20}{YQ}=\cos\theta.$$ It follows that if $XY=f(\theta)$ then $$f(\theta)=\frac{17}{\sin\theta}+\frac{20}{\cos\theta}.$$ Now it's basically all over. Differentiate. We get $$f'(\theta)=-\frac{17\cos \theta}{\sin^2\theta}+\frac{20\sin\theta}{\cos^2\theta}.$$ Set the $f'(\theta)$ equal to $0$, and do some algebra. After a short while we arrive at $$\left(\frac{\sin \theta}{\cos \theta}\right)^3=\frac{17}{20}.$$ Now solve for $\tan\theta$. We get $$\tan \theta=\left(\frac{17}{20}\right)^{1/3}.$$ Now we can find the appropriate $\theta$ to high accuracy using a calculator.

To justify that this really gives the minimum, note that there is only one one critical point. Since taking $X$ very close to $P$ or very far from $P$ is obviously bad, at that critical point we must have a minimum.

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Andre thank you a lot really ,, it was helpfull for me to understand the problem but it is a derivative problem so the solution is conected with derivations ,,anyway thank you again :) –  viola Dec 1 '12 at 19:45
The above solution uses derivatives! It is just that I left out that part, because it is easy. If it will make you feel better, I will add a couple of lines. –  André Nicolas Dec 1 '12 at 19:54
:))) thank you Andre :) –  viola Dec 1 '12 at 20:15