# 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=pir^2h
500=pir^2h
500/(pi*r^2)=h

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

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

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

2. 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 2pir^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.

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.

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.

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 r[sub]low[/sub] = 1, r[sub]high[/sub] = 500, then r[sub]lower[/sub] = 0.5 and r[sub]higher[/sub] = 501 will do.
3. Calculate the SA for each of those 4 points
4. If the point with the lowest SA is either r[sub]lower[/sub] or r[sub]higher[/sub], set r[sub]low[/sub] = r[sub]lower[/sub] and r[sub]high[/sub] = r[sub]higher[/sub], and go back to step 2.
5. Set r[sub]mid[/sub] equal to the radius that gives the lowest SA. Set r[sub]lower[/sub] to be the radius right below it and r[sub]higher[/sub] to be the radius right above it.
6. Set r[sub]low[/sub] to the midpoint between r[sub]lower[/sub] and r[sub]mid[/sub], i.e., (r[sub]lower[/sub] + r[sub]mid[/sub]) / 2. Set r[sub]high[/sub] to the midpoint between r[sub]higher[/sub] and r[sub]mid[/sub].
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 r[sub]mid[/sub] is the lowest of the 5 points, and the difference between the SA of r[sub]mid[/sub] and r[sub]low[/sub], and the difference between the SA of r[sub]mid[/sub] and r[sub]high[/sub] are both less than some tolerance, you’re done, answer = r[sub]mid[/sub]. Otherwise…
9. Find the radius with the smallest SA and go back to step 5.

Here’s what you do. Let’s call the minimum value of r, rm. So the minimum surface area is 2pirm^2 + (1000/rm). Now find the value of SA for r near rm.

SA=2pi(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 2pi(2drm+d^2)-d1000/(rm(rm+d)). If d is small enough you can throw out any factor with d^2, and eventually you get the difference is

4pidrm-1000d*(rm-d)/rm^3. Set this to zero, and cancel a bunch of stuff, and you get 1000(rm-d)=4pirm^4. But we’re interested in the situation when d=0, so 1000rm=4pirm^4 or rm=the cube root of (1000/4/pi), which is the right answer (unless I messed up)

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.

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 d[sup]2[/sup]” business, which perhaps would let us count it as a calculus-free solution.

As before, our constraint is that the total volume πr[sup]2[/sup]h = the constant V = 500. Equivalently, h = V/(πr[sup]2[/sup]). And the function to be minimized subject to this constraint is the surface area 2πr[sup]2[/sup] + 2πrh; equivalently, plugging in the value of h in terms of r as derived from the constraint, the function to be minimized is 2πr[sup]2[/sup] + 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 πr[sup]2[/sup] + V/r. Call this function A®.

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/m[sup]2[/sup] = 0, from which we derive m[sup]3[/sup] = 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/(πr[sup]2[/sup]). 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.

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.

(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?)

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.

Wait, so is there a way to solve this using algebra, without brute forcing it?

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.

(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)[sup]2[/sup] + d (by “completing the square”) reduces this to the fact that the particular function x[sup]2[/sup] 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 a[sup]2[/sup] and b[sup]2[/sup], 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 x[sup]2[/sup] is a strictly increasing function for non-negative inputs. And this is given by the observation that (x + a)[sup]2[/sup] = x[sup]2[/sup] + 2ax + a[sup]2[/sup] > x[sup]2[/sup] for positive a and non-negative x. After that, there’s really no room for anyone to plead ignorance.)