Calculus Cylinder-Cone Problem
Date: 07/13/99 at 04:53:17 From: Ted Drew Subject: Calculus Cylinder-Cone Problem (How does one "see" the simple solutions?) Based on your archives, I was able to complete the problem below, determining the relation between x and y as follows (see Maximizing the Volume of a Cylinder http://mathforum.org/library/drmath/view/53471.html ) : Find the altitide of the cylinder of maximum volume that can be inscribed in a cone having a radius of 7 and a height of 12. _ /\ | / \ | / \ | / \ | /________\ |12 /| |\ | / | | \ | / | y| \ | / | | \ | ******************* _| |-x-| |________| 7 (7X12) = 84 = area of the whole big triangle (7-x)y = combined area of the triangle on the lower sides (12-y)x = area of the top triangle The area of the square is going to be the area of the big triangle, less the areas of the little ones surrounding the square. So, 2xy = 84 - (7-x)y - (12 - y)x which, solving for y, gives you y = (84-12x)/7 We put this back into V = pix^2y and find the maximum of it using the zero point(s) of dV/dx. So that was easy. Now my real question (but first a little more). I started this problem with a different tactic for relating x and y. More specifically, looking at your same triangle diagram, I noted a Pythagorean. 7^2 + 12^2 = the hypotenus (H) of one side of the big triangle, or H = 193 _ /\ | / \ | / \ h2 | / \ | /________\ H=193 |12 /| |\ | / | | \ h1 | / | y| \ | / | | \ | ******************* _| |-x-| |________| 7 But then, one should be able to define the hypotenuse segments (h1 and h2), as follows: (7-x)^2 + y^2 = (h1)^2 x^2 + (12-y)^2 = (h2)^2 And since the hypotenuse segements add up the the hypotenuse overall, you should be able to do (h1)^2 + (h2)^2 = H = 193 [(7-x)^2 + y^2] + [x^2 + (12-y)^2] = 193 (1) So, if we _could_ isolate y out of (1), then we ought to be able to do the same plug-in operation into V = pix^2y as with the other simpler relation above, and get the same answer. Of course, I got hopelessly bogged down in trying to solve (1) for y, and because of the squares, I think you even end up with multiple solutions for y. So I just gave up. Obviously, _your_ way is a lot easier. But now for my questions: 1. Would my way be feasible, and assuming you could isolate y from (1) and plug it into the cylinder volume formula and find the maxima, would it work? ^^^^^^^^^^^^^ Is there a way of isolating y in (1) and doing it my way? 2. More philosophically, how do you know when to bail on a proposed method of solving a problem, and look for a simpler one? I am getting the idea that something is wrong with me because I did not "see" your solution for relating x and y first. I mean, if I had never come across your suggestion for relating x to y, I might have tried hopelessly and in vain with my approach for hours on end. In this case (because it was a workbook problem) I was relatively certain there was going to be a reasonable way of solving the problem. But if it had been a problem in the real world, how would I know? What guidance would I have to tell me whether to keep pursuing a difficult method or to look for a simpler one? Conversely, how would I realize that a simpler method just didn't exist, and my only recourse was to keep working away at the difficult one?
Date: 07/13/99 at 10:19:14 From: Doctor Rob Subject: Re: Calculus Cylinder-Cone Problem (How does one "see" the simple solutions?) Thanks for writing to Ask Dr. Math! Your method will work if you use the correct equations. The Pythagorean Theorem says a^2 + b^2 = c^2. You seem to be using a^2 + b^2 = c, instead. The hypotenuse should be sqrt(193), not 193, and the equation you get using your method would be sqrt(193) = h1 + h2, = sqrt((7-x)^2 + y^2) + sqrt(x^2 + (12-y)^2). If you bring one of the square roots from the right side to the left and square both sides, you will reduce the number of roots in the equation to one. Isolate that one one side of the equation and square again, and you will have a polynomial equation relating x and y. This turns out to be of the form (12*x+7*y-84)^2 = 0, so you are led back to the equation 12*x + 7*y - 84 = 0, or y = (84-12*x)/7 as we saw above. So whichever way you do the algebra, the equation relating x and y will turn out the same. The rest of the problem is identical. When you find the equation 12*x + 7*y - 84 = 0, which is the equation of a straight line, you should realize that there is a simpler way. You can identify the line as the diagonal line with negative slope in the above diagram, if you know coordinate geometry. The two-intercept form x/7 + y/12 = 1 might be the simplest way to see that. If you are working from first principles, you should say to yourself, "What if x = 0? The value of y would have to be 12 then. What if y = 0? The value of x would be 7. How does y vary with x? When x goes up by X, what happens to y?" This should lead you to the same linear equation relating x and y, again. If the problem is from a textbook, probably the methods in the section immediately preceding it will have some bearing on the solution path you choose. If the equations start to get more and more complicated, without any prospect of simplification, then you probably are on the wrong path. After all, these kind of problems are designed to be exercises to give you experience in using the methods of that section, and shouldn't be too terrible in terms of the burden of computation. If the problem arises in some other context, then you can't have that kind of confidence when the calculations start to get messy. In general, when faced with triangles, I try to use similarity and its accompanying proportionality first. Then I try using the areas. Finally I try the Pythagorean Theorem. I use this order because the Pythagorean Theorem usually gives me a square root, which is harder to deal with than the equations from the other two methods. Sometimes it is the best way, or even the only way, but I try the other two attacks first. I may even try starting all three, more or less in parallel, and see which one looks the most promising after a few steps. I hope this helps! If not, write again. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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