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#1
03-05-2010, 07:10 PM
 brickbacon Guest Join Date: Nov 2004
Math Question

Two questions actually.

If you are trying to minimize the surface area of a cylinder with a given volume (500), is there better way to solve this than the following:

1. Solve for height in terms of radius using the equation for volume

V=pi*r^2*h
500=pi*r^2*h
500/(pi*r^2)=h

2. Substitute that value for height in the surface area equation.

SA= 2*pi*r^2 + 2*pi*r*h
SA= 2*pi*r^2 + 2*pi*r*(500/(pi*r^2)
SA= 2*pi*r^2 + (1000/r)

3. Then you graph the last formula, and trace the line to find the y-minimum.

4. Then plug in the r value at the y-min to the volume equation to find the height.

Ok, now I know that will eventually give you the correct answer if you are discerning enough to find the minimum correctly, but is there another way to solve this without calculus. I know you can set the derivative to 0, but my students do not know calculus yet.

Also, if a question specifies that the top and bottom portions of the can (the parts represented in the SA equation by 2*pi*r^2) cost twice as much as the sides, would you just double that portion of the equation when graphing it to figure out the proper height/radius? Thanks in advance.
#2
03-05-2010, 07:25 PM
 CalMeacham Guest Join Date: May 2000
You can always do what Descartes did, and plug in a range of values, find the loest two of these, and continually interpolate, plugging in values and calculating until you get a value that changes by less than some acceptable increment. Solution by brute force.

You could also use something like Newton's Method for solving for a minimum, but that's still using calculus, albeit in disguise.

http://en.wikipedia.org/wiki/Newton's_method

Last edited by CalMeacham; 03-05-2010 at 07:25 PM.
#3
03-05-2010, 07:55 PM
 Hari Seldon Guest Join Date: Mar 2002
Sometimes there are clever ways to do these things without calculus. For example, to find the largest rectangle with perimeter P, let x and y be the sides. Then the area is P^2/16 - (x-y)^2/4 and it is then obvious that x = y gives the max. But I certainly don't know anything involving cylinders, although such a trick probably exists, but I am not about to waste time finding it when I can do it so easily using calculus. In fact this sort of thing is exactly why calculus is important. It obviates the need for such special tricks.
#4
03-05-2010, 08:24 PM
 Punoqllads Charter Member Join Date: Jul 2000 Location: Silly Cone Valley, CA Posts: 3,078
1) Pick two radius values that you believe bound the correct solution.
2) Pick a radius value further out than both of those two points. E.g., if you've picked rlow = 1, rhigh = 500, then rlower = 0.5 and rhigher = 501 will do.
3) Calculate the SA for each of those 4 points
4) If the point with the lowest SA is either rlower or rhigher, set rlow = rlower and rhigh = rhigher, and go back to step 2.
5) Set rmid equal to the radius that gives the lowest SA. Set rlower to be the radius right below it and rhigher to be the radius right above it.
6) Set rlow to the midpoint between rlower and rmid, i.e., (rlower + rmid) / 2. Set rhigh to the midpoint between rhigher and rmid.
7) Calculate the SA for the new 2 points. You've already calculated the SA of the other 3 points.
8) If the SA of rmid is the lowest of the 5 points, and the difference between the SA of rmid and rlow, and the difference between the SA of rmid and rhigh are both less than some tolerance, you're done, answer = rmid. Otherwise...
9) Find the radius with the smallest SA and go back to step 5.
#5
03-05-2010, 10:57 PM
 Andy L Member Join Date: Oct 2000 Posts: 2,306
Quote:
 Originally Posted by brickbacon Two questions actually. If you are trying to minimize the surface area of a cylinder with a given volume (500), is there better way to solve this than the following: 1. Solve for height in terms of radius using the equation for volume V=pi*r^2*h 500=pi*r^2*h 500/(pi*r^2)=h 2. Substitute that value for height in the surface area equation. SA= 2*pi*r^2 + 2*pi*r*h SA= 2*pi*r^2 + 2*pi*r*(500/(pi*r^2) SA= 2*pi*r^2 + (1000/r) 3. Then you graph the last formula, and trace the line to find the y-minimum. 4. Then plug in the r value at the y-min to the volume equation to find the height. Ok, now I know that will eventually give you the correct answer if you are discerning enough to find the minimum correctly, but is there another way to solve this without calculus. I know you can set the derivative to 0, but my students do not know calculus yet. Also, if a question specifies that the top and bottom portions of the can (the parts represented in the SA equation by 2*pi*r^2) cost twice as much as the sides, would you just double that portion of the equation when graphing it to figure out the proper height/radius? Thanks in advance.
Here's what you do. Let's call the minimum value of r, rm. So the minimum surface area is 2*pi*rm^2 + (1000/rm). Now find the value of SA for r near rm.

SA=2*pi*(rm+d)^2 + (1000/(r+d)) where d is the distance from rm to r.

The difference between the minimum surface area and the surface area with r=rm+d is 2*pi*(2*d*rm+d^2)-d*1000/(rm*(rm+d)). If d is small enough you can throw out any factor with d^2, and eventually you get the difference is

4*pi*d*rm-1000*d*(rm-d)/rm^3. Set this to zero, and cancel a bunch of stuff, and you get 1000(rm-d)=4*pi*rm^4. But we're interested in the situation when d=0, so 1000rm=4*pi*rm^4 or rm=the cube root of (1000/4/pi), which is the right answer (unless I messed up)
#6
03-06-2010, 12:53 AM
 Chronos Charter Member Join Date: Jan 2000 Location: The Land of Cleves Posts: 50,877
That works, Andy L, but you're really using calculus there, you're just deriving the pieces of it you need from scratch, so it doesn't look like you're using calculus.
#7
03-06-2010, 08:41 AM
 Andy L Member Join Date: Oct 2000 Posts: 2,306
Quote:
 Originally Posted by Chronos That works, Andy L, but you're really using calculus there, you're just deriving the pieces of it you need from scratch, so it doesn't look like you're using calculus.
Very true. That could be a good thing, since it introduces the students to the usefulness of calculus concepts without needing to introduce the more conceptually difficult pieces of it (differentials, etc.) - if these students are eventually going to get to calculus, my method might serve as a stepping stone.
#8
03-06-2010, 02:59 PM
 Indistinguishable Guest Join Date: Apr 2007
I agree with Hari, Chronos, and Andy L that this is perhaps a good opportunity and method by which to actually start motivating and introducing some of the ideas of calculus. But, all the same, we could understand Andy's method without any of the "If d is small enough, you can toss out d2" business, which perhaps would let us count it as a calculus-free solution.

As before, our constraint is that the total volume πr2h = the constant V = 500. Equivalently, h = V/(πr2). And the function to be minimized subject to this constraint is the surface area 2πr2 + 2πrh; equivalently, plugging in the value of h in terms of r as derived from the constraint, the function to be minimized is 2πr2 + 2V/r, where we no longer have to worry about the constraint. And, of course, we can factor out the 2, so all we really want to minimize is πr2 + V/r. Call this function A(r).

Now, observe that A(x) - A(m) = (π(x + m) - V/(xm))(x - m). Suppose we want m to be the desired r uniquely minimizing A; that's the same as saying this expression should always be positive for x distinct from m. Of course, if x > m, then x - m is positive, so we will require that for x > m that π(x + m) - V/(xm) be positive as well. And for x < m, x - m is negative, so we will require that for x < m that π(x + m) - V/(xm) be negative as well. But this means π(x + m) - V/(xm) must switch over from positive to negative as x switches from above to below m; accordingly, by continuity, when x = m, we must have that π(x + m) - V/(xm) = 0. That is to say, 2πm - V/m2 = 0, from which we derive m3 = V/(2π).

Thus, the uniquely minimizing radius is the cube root of V/(2π) (in this case, the cube root of 500/(2π)), with the corresponding height determined by h = V/(πr2). In this method, of course, calculus was lurking at every corner, sometimes dangerously close to letting itself out, but perhaps we could be said to have avoided actually allowing it to break through, never having actually invoked any infinitesimals or assumptions of local linearity...

Last edited by Indistinguishable; 03-06-2010 at 03:03 PM.
#9
03-06-2010, 03:24 PM
 Chronos Charter Member Join Date: Jan 2000 Location: The Land of Cleves Posts: 50,877
Quote:
 In this method, of course, calculus was lurking at every corner, sometimes dangerously close to letting itself out, but perhaps we could be said to have avoided actually allowing it to break through, never having actually invoked any infinitesimals or assumptions of local linearity...
The pedant in me does feel obligated to point out that you did invoke an assumption of continuity, but since students won't learn that that has anything to do with calculus until long, long after they've taken calculus (if ever), that's probably OK.
#10
03-06-2010, 03:31 PM
 Indistinguishable Guest Join Date: Apr 2007
Yes, I thought about that. Fair point, in that really, continuity itself contains the seeds of all the notions of limits, infinitesimals, derivatives, and so on already. That having been said, the particular fact that "Nice functions (e.g., linear combinations of integer powers of x) can only cross from positive to negative by going through critical points where they are 0 or undefined" (i.e., nice functions satisfy the intermediate value property, which is strictly weaker than formal continuity) is one that I think almost all students are familiar with well before officially learning calculus. It is, after all, the same principle which leads us to conclude that all positive numbers actually have square roots and so on ("Because we switch over at some point from having a lower square to a higher square, surely there is a crossing point"), even if more rigorous exploration of it is a topic not actually touched till Analysis.

Last edited by Indistinguishable; 03-06-2010 at 03:35 PM.
#11
03-06-2010, 03:44 PM
 Indistinguishable Guest Join Date: Apr 2007
(Is it really true, though, that students won't learn that continuity has anything to do with calculus till long after? I mean, my own experience with calculus was nonstandard and largely self-taught starting from the motivating idea of continuity, but I did experience the first few weeks of an intro high school calc course, and I seem to recall there being an initial emphasis on introducing the idea of limits, and of continuity as preservation of limits (which is very close, though I don't recall the point being made explicitly, to the converse view of limits as values given by continuity), and on then using limits to define derivatives, and so on, forging the bridges between these notions. Is this material not standard in calculus education? Or is it that it is merely perfunctory and largely forgotten later on without explicit return to it, just as, say, delta-epsilon definitions?)

Last edited by Indistinguishable; 03-06-2010 at 03:46 PM.
#12
03-06-2010, 04:48 PM
 Chronos Charter Member Join Date: Jan 2000 Location: The Land of Cleves Posts: 50,877
Even with me being in a technical field, and taking a lot of math courses, I don't think I ever encountered a rigorous definition of continuity until grad school. Before then, it was mostly just considered as obvious, or at best, the intermediate value "theorem" being used as a definition of continuity.

Last edited by Chronos; 03-06-2010 at 04:49 PM.
#13
03-06-2010, 05:24 PM
 Bayesian Empirimancer Guest Join Date: Mar 2010
Wait, so is there a way to solve this using algebra, without brute forcing it?
#14
03-07-2010, 01:57 AM
 Indistinguishable Guest Join Date: Apr 2007
Quote:
 Originally Posted by Chronos Before then, it was mostly just considered as obvious, or at best, the intermediate value "theorem" being used as a definition of continuity.
Ah, alright. Though, to be sure, the intermediate value property certainly yields a fine definition of a useful concept that could be given the ordinary language name of "continuity", just not the particular one that history happened to bestow that name upon in official mathematical jargon. But, at any rate, it's all we need here.

Bayesian Empirimancer: Sure. See post #8.

Last edited by Indistinguishable; 03-07-2010 at 02:02 AM.
#15
03-07-2010, 03:15 AM
 Indistinguishable Guest Join Date: Apr 2007
Quote:
 Originally Posted by Indistinguishable But, at any rate, it's all we need here.
(In case there is still any concern by anyone over the invocation of the stronger-than-necessary condition of "continuity" in post #8, let me it put another way: the function in question is a rational function, and thus the only niceness condition we need to know is that the output of a polynomial can't switch sign without passing through a zero along the way. Even without raising the spectre of continuity, this is surely a basic precalculus fact/assumption (actually, one essentially equivalent to the fundamental theorem of algebra, though probably most students simply take it as given and are not exposed to such connections).

In fact, even if they aren't aware of even this basic fact in general, the relevant function is π(x + m) - V/(xm). We can assume x is positive, so the sign of this function is unaltered by multiplying it by x, which results in a quadratic function. So all that is needed is the fact that quadratic functions satisfy the intermediate value property. And the well-known decomposition of an arbitrary quadratic polynomial into the form k(x - r)2 + d (by "completing the square") reduces this to the fact that the particular function x2 satisfies the intermediate value property (as all other quadratics are mere translations of scalings of this). That is, the only needed fact is that if s is inbetween a2 and b2, then s has a square root inbetween a and b. Splitting into cases according as to the signs of a and b, this readily follows from the fact that x2 is a strictly increasing function for non-negative inputs. And this is given by the observation that (x + a)2 = x2 + 2ax + a2 > x2 for positive a and non-negative x. After that, there's really no room for anyone to plead ignorance.)

Last edited by Indistinguishable; 03-07-2010 at 03:19 AM.

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