Solve Calculus Problem: Minimal Time to Travel 12 km Down Shoreline

[Jeebus]

I was in my Calculus class and a problem came up that I couldn't figure out. Our teacher wrote down a problem on the board to test us on the upcoming exam next week and he called upon us randomly, but no one got it. I thought I figured it out, but I was off. Hopefully you can help me here.

A person on a boat in a lake is 9 km from the shore and must go to a
point 12 km down the shoreline in the shortest possible time. The
person can walk 8 km per hour and the boat can travel r km per hour.

a) Assume that the person should travel by boat and by foot. Let d be
the distance down the shoreline the person should strike land for the
shortest total travel time. Write d as a function of r.

b) Sketch the graph of the function d(r). Determine the slowest speed
of the boat so the shortest possible time criterion is met by making
the entire trip by water. Use this to give the domain of the function
in the context of this problem.

c) Determine the concavity of the graph of the function d over the
domain stated in part b. What information does the concavity give about the relation between d and r?
Our class attempted this problem and wrote an equation based on the
given information. We took the derivative of our equation to find the
answer to part a, but the answer we arrived at was different from the
answer given in the book. Our answer was d=(r-36)/4, from the equation d(r) = r * sqrt(d^2+81) + 8(12-d). As stated, d is the distance down the shoreline where the person should strike land. Thus 12-d is the distance from the point where the person lands to his or her destination. Using the Pythagorean Theorem, we found the distance from the boat to the point where the person will land to be the square root of (d^2 + 81). The answer the book gave us is d(r) = 9r/sqrt(64-r^2).

I, (and the class can't figure out) or understand what we did wrong, and I hope someone will be able to help me for tomorrow since no one figured it out. Some got close, but no one got it completely.

Here is what I think:

I believe that we were writing d in terms of rate, not rate and time, so we need to get rid of the time component, right? Then time for the trip is t = sqrt(d^2+81)/r + (12-d)/8, right? So then you would differentiate it?

But then I get lost. :)

 

[NateTG]

$$\frac{\sqrt{d^2+81}}{r}+\frac{12-d}{8}$$

Now, the derivative with respect to $$d$$:

The derivative of the second term is clearly:
$$-\frac{1}{8}$$
So that leaves the first term.
$$\frac{d}{dd}\frac{(d^2+81)^{\frac{1}{2}}}{r}$$
If you apply the chain rule, you get:
$$\frac{1}{r} 2d \frac{-1}{\sqrt{d^2+81}}$$

So now, you need to solve:
$$0=\frac{-2d}{\sqrt{d^2+81}}-\frac{1}{8}$$

Then you can check those solutions, and the endpoints to see which is the optimal solution

 

[Jeebus]

Thank you, NateTG I really appreciate you showing the steps. I figured out what I was doing wrong.

Thanks again.