Optimization (Time/Distance Equation)

So I finally figured out the last question on the packet we had in class, I figured it out the morning before our Friday class, and when the professor went over it I was happy to see I was right, but I had to go to the book for help, which is the first time ever I have had to use the book (other than homework).

The equation involved a man throwing a stick for his dog into a lake, and then figuring out how long he had to run on land, and then swim in the water to get their the fastest/most efficiently.

Today I am going to show a similar problem, and how it would be solved.

Example: A man is taken hostage in a vehicle driving along a desert road. The man knows the desert well, because he (being wealthy and very paranoid) built a hidden apocalyptic bomb shelter/hideout along this very road. The man realizes that his only option for escape is to wait for an opportunity to open the car, roll out and sprint to his shelter. The vehicle is traveling at 60 miles per hour, The man knows he can sprint at exactly 9 miles per hour (he’s been marathon training at this speed, so he can maintain it for well over an hour). How long does the man need to stay in the car before he jumps out in order to make it to his secret shelter in the shortest (thus safest) amount of time? (For our purposes pretend the rolling takes such a minimal amount of time there is not a need to factor in time lost for it).

Here’s a nice diagram for visual aid (and additional info the man already knew.. like distance) :

How to Solve:

1. First we need to figure out the length he’ll be running in the desert. In order to do this, we can use Pythagorean’s theorem, and we get the following equation:

$L=\sqrt{10^2+x^2}$ where L stands for the length he’ll be running.

Now that we have that, we can create a time function.

2. If distance = rate x time, Time is equal to the distance over the rate.

If $d=rt$ , then $t=\frac{d}{r}$

We can use this to make a function for our problem.

3. $T=\dfrac{15-x}{60}+\dfrac{\sqrt{10^2+x^2}}{9}$

4. Now find it’s derivative: $T'=\frac{-1}{60}+\dfrac{x}{9\sqrt{10^2+x^2}}$

and set it equal to zero.

5. Solve for X (you may need to use the quadratic equation on other problems like this). and you get:

$\dfrac{x}{9\sqrt{10^2+x^2}}=\frac{1}{60}$

cross multiply

$60x=9\sqrt{10^2+x^2}$

square each side

$3600x^2=81(10^2+x^2)$

$3600x^2=8100+81x^2$

$3519x^2=8100$

$x^2=2.30179$

$x=1.5$ (approximately)

6. Now that we know what x is, we can solve for distance driving and distance running.

Distance Driving = 15-x = 13.5

Distance Running = $\sqrt{10^2+x^2}$ which is aprx 10.1 miles

7. How long will it take him? Let’s plug it in:

$T=\dfrac{13.5}{60}+{10.1}{9}$ $T=1.34$ hours. Let’s hope he can evade them for that long. What if he waited till the last minute and jumped out at 15 miles?

$T=\dfrac{15}{60}+{10}{9}$ $T=1.36$ That’s worse… lets say he jumped out at our starting point of 0 miles? Which would be apx 18 miles away (the long side of the triangle) $T=\dfrac{18}{9}$ $T=2$ . That would take him 2 hours!

Obviously we can see these in a similar way we see end points and critical points. We had to solve for the derivative to find the critical point, which in this case, was our poor character’s best chance at survival, and the first and last points on the map (aka the triangle) were the end points.