Where to Land on the Beach?Date: 5/22/96 at 16:26:29 From: Anonymous Subject: Calculus An island is at point A, 6 miles offshore from the nearest point B on a straight beach. A store is at point C, 7 miles down the beach from B. If a person can row at the rate of 4 mi/hr and walk at a rate of 5 mi/hr, where should the person land in order to go from the island to the store in the least possible time? Date: 5/31/96 at 16:58:8 From: Doctor Anthony Subject: Re: Calculus With the problem as given, the best course of action is to row straight from A to C. If you land on the beach between B and C you will take longer. I suspect that the numbers in the question should be changed to give a bigger advantage to walking on the beach, but I will work through the calculus anyway to show the method. Let x be the distance from B alng the beach where the person lands. The distance he rows is then sqrt(x^2 + 36) and the rowing time is sqrt(x^2+36)/4. The distance he walks will be (7-x), and time walking is (7-x)/5. Total time = sqrt(x^2+36)/4 + (7-x)/5 Now differentiate with respect to x. dt/dx = (1/4)(1/2)(1/sqrt(x^2+36))(2x) - (1/5) = 0 for max. or min. (x/4)(1/sqrt(x^2+36)) = (1/5) 5x = 4sqrt(x^2+36) Square the equation 25x^2 = 16(x^2+36) 25x^2 = 16x^2 + 576 9x^2 = 576 And taking square roots 3x = 24 x = 8 And we see that this means landing beyond point C to obtain a theoretical minimum turning point. In practice we should row directly to C. I feel sure the numbers given in the question are not those intended, but the method to use is exactly as outlined above. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/07/2003 at 05:06:01 From: Ajit Subject: Calculus/Maxima In the above problem, Dr. Anthony says that in real life you would row directly to point C. Why is your calculus at variance with the obvious answer of going directly to to point C? 5x = 4sqrt(x^2+36) Square the equation Now look at this equation carefully. Coefficients 4 and 5 are really the speed of rowing and walking respectively. Correct? Consider the situation where the two speeds are equal. Would you then get (Speed of walking)x = (speed of rowing)sqrt(x^2+36) and hence x = sqrt(x^2+36) which is clearly unacceptable? This is precisely why with speeds as given viz. 4 mph and 5 mph Dr. Anthony has arrived at x = 8 miles. The whole analysis seems to be erroneous. Would you please explain? A cursory glance at the situation tells us that: If the speed of rowing >> speed of walking then one should land close to B to minimize the time taken. If speed of rowing << speed of walking then go close to C. If the two speeds are equal then go directly to C. In any other situation the landing point must be between B and C. Your analysis does not reflect this at all. Why? Date: 08/07/2003 at 12:58:16 From: Doctor Peterson Subject: Re: Calculus/Maxima Hi, Ajit. Let's deal with two issues: first, why the calculus yields a result that has to be overridden; and second, why your "cursory" expectations are wrong. When calculus is applied to an optimization problem, we have to consider not only the equations, but also constraints. In this case, it was assumed when the equations were written that the landing point was on the segment BC, so that x<7. If this were not done, then the equation would have needed an absolute value: Total time = sqrt(x^2+36)/4 + |7-x|/5 This was not done, in order to simplify the work, since it is obvious that x must be between 0 and 7 for a valid solution. But it means that the constraint on the equation has to be checked. When it was found that the optimal value of x was 8, outside the bounds of the equation, this meant that the true optimum had to be found at a boundary point. If the absolute value formulation had been used (which in effect combines this equation, with its constraint, with another equation that is applied in a different region, to make a piecewise-defined function), then it would have been found that neither function has a minimum, and again the boundary between the functions would have to be checked. The true optimum would be not at a place where the derivative is zero, but where the derivative does not exist. (You'll see below what happens when the two speeds are the same, which causes a slightly different problem.) Now, let's consider your claims: >A cursory glance at the situation tells us that: >If the speed of rowing >> speed of walking then one should land close >to B to minimize the time taken. >If speed of rowing << speed of walking then go close to C. >If the two speeds are equal then go directly to C. >In any other situation the landing point must be between B and C. >Your analysis does not reflect this at all. Why? Sometimes what is "obvious" is true; sometimes not. Your claims don't really make sense to me; perhaps you wrote it wrong. If I row fast, I would expect to spend all or most of my time rowing, so I would land near C; if I walk faster, I would want to do more walking, by landing closer to B. I would never expect to land at B exactly, since just a small amount of extra rowing would save a good bit of walking. And I wouldn't have to do both at the same speed in order to make a direct route to C a good idea. I would expect that there would be some minimum ratio of rowing to walking speed that would make C the best landing point. Let's see under what conditions the landing point is actually between B and C. I'll replace the speeds 4 and 5 with u and v, and the distances 6 and 7 with a and b; then we have t = sqrt(x^2+a^2)/u + (b-x)/v The derivative is then dt/dx = (1/u)(1/2)(1/sqrt(x^2+a^2))(2x) - (1/v) and when we set this to zero we get x = ua/sqrt(v^2 - u^2) If u >= v, this is undefined. In that case, we have to compare the times when x=0 and when x=b; these are t(0) = a/u + b/v t(b) = sqrt(b^2 + a^2)/u It can be shown that if u >= v, then t(0) > t(b), so the optimal strategy is to go directly to C. That agrees with my version of your ideas. The other extreme will be when the optimal x is >= b, as in our original problem. Then again the optimal strategy will be to go to C. When does this happen? When ua/sqrt(v^2 - u^2) >= b This can be solved to give u/v >= b/sqrt(a^2 + b^2) That is, when the ratio of rowing to walking speeds is at least that of the maximal walking distance to the maximal rowing distance, it will be better to row all the way. (Note that the right side here is less than 1, so that this case includes the case where u >= v.) So it turns out that the optimal landing point will be between B and C when u/v < b/sqrt(a^2 + b^2) And it will NEVER be best to land at B (unless, of course, b=0 so that B=C; or it a or u is 0, which don't make sense). I may well have made some mistakes here or there, but since the result agrees with my intuition, it "must" be correct ;-) If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 08/07/2003 at 22:05:11 From: Ajit Subject: Thank you (Calculus/Maxima) Thanks a lot Dr Peterson. I obviously got confused between points B and C in my "thoughts". But now you have cleared my doubts and thanks are due to you and the Math Forum. |
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