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?
 A: 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. 
