Topic 5.10: Introduction to Optimization Problems

Elvis

More than twenty years ago, on a beach on Lake Michigan, a math professor named Tim Pennings used to play fetch with his Welsh corgi, Elvis. He’d stand at the water’s edge and throw a tennis ball out into the lake, and Elvis would tear off after it. Pennings noticed something odd about how Elvis did it. He’d run along the sand for a while, then plunge in and swim the rest of the way at an angle.

He could have jumped in right away and swum straight at the ball, the shortest path, or run until he was level with it and swum straight out, the least swimming. Almost every time, he picked something in between.

Pennings wondered whether his dog was picking the fastest route, so with a friend and a stopwatch, he timed him. On the sand Elvis ran more than six meters every second, and in the water less than one, so he was about seven times faster on land.

By the end of this lesson, you’ll find the exact spot where Elvis should jump in. You’ll prove that nothing on this beach beats it, and then you’ll check his real choices against yours. His real numbers are a little messy, so let’s start on a friendlier beach, with an Elvis who swims a lot better than the real one ever did.

The shortest path loses

Our made-up dog runs five meters a second on the sand and swims three, which is faster than world-record pace for the fifty-meter freestyle, but it makes the numbers come out beautifully. The ball lands four meters out, and the point on the shore closest to it, let’s call it C, is twelve meters down the beach from where he starts.

Before we time anything, I want your gut call. Of these three routes, which one do you think gets him to the ball first?

Let’s race all three at once. One ghost is in the water right away, one is flying along the sand toward C, and the middle one cuts in at an angle… …and the middle route wins, in about three point six seconds. Running to C comes second at three point seven three, and the shortest path of all comes in dead last, at four point two two.

The shortest path loses because all of it is in the water, where he’s slow, and running all the way to C wastes time on the sand. Somewhere in between, there’s one best place to jump in.

Now you find it. Drag the screwdriver along the tape to choose where he jumps in, and send him off as many times as you like. Every run drops a dot on the time graph.

Your dots say about three meters before C, even faster than the middle route from the race. But is it exactly three, or two point nine, and even after a thousand tries, how could you ever be sure that nothing between them is faster? Calculus can find the bottom of that valley exactly and prove it, so let’s turn the beach into a function.

Turn the beach into a function

Let’s draw the beach from above. Elvis starts at A, the ball is four meters out from C, and C is twelve meters down the beach. The one thing Elvis chooses is where he jumps in, so let x be how many meters before C he goes into the water. He runs the rest of the beach first, twelve minus x meters.

His swim is the long side of a right triangle with legs x and four, so by Pythagoras it’s the square root of x squared plus sixteen. Time is distance divided by speed, so the run takes twelve minus x over five seconds, the swim takes that square root over three, and T is the total.

x equals zero means he runs all the way to C, and x equals twelve means he jumps in at A and swims the whole way. Every choice he has sits on the closed interval from zero to twelve.

We’ve just built an optimization problem. We’re looking for the input that makes a quantity as small or as large as it can be, and here that quantity is the total time.

Where the slope goes flat

The function fills in every entry point you didn’t try, and its lowest point still looks close to three. At the bottom of the valley the tangent line goes perfectly flat, sloping down before it and up after it. A flat tangent means T prime equals zero, and that makes the bottom a critical point we can find exactly. The running part is twelve fifths minus x over five, so its derivative is just minus one fifth.

To differentiate the square root, I’ll write it as a one-half power. The outer derivative brings down a half and makes the exponent minus a half, then we multiply by two x, the derivative of the inside, and that two cancels the half. We now have T prime, in seconds per meter. The minus one fifth comes from the running time and the fraction comes from the swimming time, so let’s go back to the beach and see what those two rates are doing.

Say Elvis jumps in one meter before C, and we nudge his entry point a little farther from C. Every meter of sand he skips saves a fifth of a second, and that’s the minus one fifth.

His swim grows much more slowly, because he’s swimming almost straight out and the nudge is sideways to that. Right here it costs only about eight hundredths of a second per meter, and that’s the second piece.

At x equals one, T prime is about minus point one two seconds per meter. What does that tell Elvis to do? As the entry point slides farther out, the swim’s cost per meter keeps climbing, and by six meters it’s bigger than the running savings.

At three meters before C the bars match exactly. The swim’s cost has grown to a fifth of a second per meter, the same as the running he saves, and that balance is what T prime equals zero means.

At the best spot, the time saved and the time added balance. So we set T prime equal to zero and add a fifth to both sides. Clearing the fractions leaves five x equals three times the square root of x squared plus sixteen. Squaring both sides gives twenty-five x squared equals nine x squared plus one forty-four, so sixteen x squared is one forty-four, and x squared is nine.

That’s three or minus three. Minus three is off our interval, and it doesn’t even solve the equation before we squared it, because five x would be negative and the root can’t be. Squaring sneaked it in.

A critical point is only a candidate, though, so let’s compare T there and at both ends of the interval. Running all the way to C takes twelve fifths plus four thirds, about three point seven three seconds, and swimming the whole way takes about four point two two.

Now you finish it off. Using the calculator if you like, what’s the least possible time for the trip, to the nearest thousandth of a second?

That means Elvis should jump in three meters before C, after running nine, and the least possible time is fifty-two fifteenths of a second, about three point four six seven. Those are two different numbers answering two different questions. Three meters tells you where, and about three and a half seconds tells you how long, so read the question twice.

On the exam you also have to justify the minimum, and a sentence like this one earns the point. T is continuous on the closed interval from zero to twelve, and its only critical point there is three.

T of three is smaller than T at both ends, so the minimum time is about three point four six seven seconds. Signs work too, since T prime is negative before three and positive after it. Just saying T prime equals zero won’t earn it, because a flat tangent can also sit at a maximum, or at no extreme at all, like x cubed at zero.

Move the ball

Look at the equation we solved one more way. Multiply both sides by three, and it says x over the square root of x squared plus sixteen equals three fifths. On the triangle that’s the sideways part of the swim over the whole swim, and three fifths is his swim speed over his run speed. That’s why the best swim came out as a perfect three-four-five triangle. The two speeds alone decide the shape of the swim.

Let me give you a puzzle. If the ball lands twice as far down the beach, twenty-four meters, but still four meters out, where’s the best place to jump in now?

Here’s the best entry point for that throw. It’s three meters before C again, with the swim at exactly the same angle. Now try a short throw. Two meters down the beach, still four meters out.

With a throw that short… where should he jump in? The math’s favorite spot is a meter behind him, outside the stretch from A to C. Backing up first would only add running and swimming, so the best choice is the end of the interval, and he jumps straight in.

Now the interval runs from zero to two, and three isn’t on it. T prime hasn’t changed, because the beach length vanished when we differentiated. It’s negative across the whole interval, so the time keeps dropping right to the end. That’s why we always check the endpoints.

So jumping straight in can be the right call. A few times Elvis did it when it wasn’t, and Pennings left those throws out, joking that even an “A” student can have a bad day.