Phil Quiz for Wednesday

[merjet]

Yes, Merlin also made an assumption, which I think is necessary to use differential calculus to minimize the surface function. Without an assumption you have 3 unknowns and 2 equations.

Agreed. The same method could be used for the "with top" situation, but the surface equation would be surface = 2*width*width + 4*height*width.

[Philip Coates]

Merlin's write up of this is a model of economy - very short and to the point. Good job!! (The assumption is valid because you can adjust length and width in proportion without changing the overall surface area.)

> Merlin also made an assumption, which I think is necessary to use differential calculus to minimize the surface function. Without an assumption you have 3 unknowns and 2 equations.

Not true that you can't use calculus to solve it without this assumption, GS. See below.

The only thing a person who doesn't remember his h.s. calculus section on maxima and minima might not remember is that the first derivative of a function = 0 is a a test for where the function has a (local) maximum or minimum and obviously in this case it has to be a minimum (because as I think Brant pointed out, you could 'stretch' the thing out along an axis and make the materials much greater.) Also you could use the second derivative to tell whether it's a maximum or minimum.

You -can- solve the problem as an equation of the form GS mentions which reduces to one equation with two unknowns: Area = (512/y) + 2xy + (512/x). But then you are doing multi-variable calculus (on a 3-D surface not on a curve) and that is -definitely- not senior year of h.s. level calculus, which is why I wasn't thinking of it as a problem that could be solved by those on the list who are not "math guys" but who recognize that it's a simple maxima-minima problem and remember that part of their schooling:

Merlin's ingenious assumption hadn't occurred to me because I was viewing the problem as an exercise in calculus with a function of two variables, not trying to make it simpler.

,,,,

I'm glad we did this. One of the cool things about math is how you can often reduce a hard problem to a simpler one. [i tutored differential equations to a chemistry grad student a few years ago, and one technique that kept cropping up was to solve a special case and then 'extend' that to solve more general, more difficult problems.]

Also one of the nice things about this kind of problem is it's very -real world-. You constantly have to find ways to make minimum use of resources, and not just in packaging.

Edited August 19, 2010 by Philip Coates

[merjet]

Merlin's write up of this is a model of economy - very short and to the point. Good job!! (The assumption is valid because you can adjust length and width in proportion without changing the overall surface area.)

Thanks, but I didn't believe it was that hard from the start. It only requires some basic calculus knowledge. Or Excel Solver!

I agree with your last sentence. However, the volume decreases, which violates the problem definition.

[Philip Coates]

No, I meant you can adjust the length in width so the volume doesn't change - increase one, decrease the other so their product remains the same.

[tjohnson]

Out of curiosity I checked into this and it turns out (as Phil stated) that you can indeed use partial derivatives to help solve this.

For functions of more than one variable, similar conditions apply. For example, in the (enlargeable) figure at the right, the necessary conditions for a local maximum are similar to those of a function with only one variable. The first partial derivatives as to z (the variable to be maximized) are zero at the maximum (the glowing dot on top in the figure). The second partial derivatives are negative. These are only necessary, not sufficient, conditions for a local maximum because of the possibility of a saddle point. For use of these conditions to solve for a maximum, the function z must also be differentiable throughout. The second partial derivative test can help classify the point as a relative maximum or relative minimum.

Using the equation S = xy + 512/y + 512/x and taking partial derivatives with respect to x and y and setting to zero, we find that this only happens when x = y = 8. Then z = 256/64 = 4.

[merjet]

Some readers might wonder how Phil and general semanticist got S = xy + 512/y + 512/x.

Let x = length, y = width, z = height, S = surface, V = volume.

For the tank w/o a top: S = 2xz + 2yz + xy and V = xyz = 256.

From the latter z = 256/(xy). By substitution in the former:

S = 2x(256/(xy)) + 2y(256/(xy)) + xy, which simplifies to:

S = 512/y + 512/x + xy

[Philip Coates]

GS and MJ, I'm curious did you guys have a career in a math or engineering related field?

I suspect someone without that kind of background wouldn't even know what a partial derivative is, for example. Non-math people might think it is differentiation done by a lazy student who doesn't finish what he starts.

[merjet]

I was an actuary (FSA). For my Bachelor of Science, I majored in math and took some pretty advanced math classes, e.g. "diffy-q", abstract algebra (groups and rings), real analysis, topology. For some of those, the students were mostly or nearly all grad students.

I suspect someone without that kind of background wouldn't even know what a partial derivative is, for example. Non-math people might think it is differentiation done by a lazy student who doesn't finish what he starts.

Nah. It's the substance that comes from scraping a partial denture.

Edited August 20, 2010 by Merlin Jetton

[tjohnson]

GS and MJ, I'm curious did you guys have a career in a math or engineering related field?

I suspect someone without that kind of background wouldn't even know what a partial derivative is, for example. Non-math people might think it is differentiation done by a lazy student who doesn't finish what he starts.

LOL, you mean like my partial denture? I did my degree in Honours Mathematics with a lot of Physics thrown in for good measure. My career, however, has been mostly in surveying and computer programming. And you?

[Philip Coates]

I got a Bachelor's in Math with a minor in Physics, then I got a Master's in Math with a minor in Philosophy. I became a computer programmer/systems analyst/systems engineer/business as a first career.

Then I "turned to the Dark Side" and went into the humanities, teaching mostly History, Literature, and Composition as a second career. (But I also make sure not to lose the math and science by tutoring or teaching some math and physics every few years.)

"abstract algebra (groups and rings).. topology" -- Those were my favorite subjects in grad school. Both of them are totally forgotten now, never having used them as neither is a practical subject.

Edited August 20, 2010 by Philip Coates

[Philip Coates]

I've always liked being a generalist rather than a specialist over the course of a lifetime. But one problem is retention in very disparate areas unless you can link them somehow.

Edited August 20, 2010 by Philip Coates

[BaalChatzaf]

I got a Bachelor's in Math with a minor in Physics, then I got a Master's in Math with a minor in Philosophy. I became a computer programmer/systems analyst/systems engineer/business as a first career.

Then I "turned to the Dark Side" and went into the humanities, teaching mostly History, Literature, and Composition as a second career. (But I also make sure not to lose the math and science by tutoring or teaching some math and physics every few years.)

"abstract algebra (groups and rings).. topology" -- Those were my favorite subjects in grad school. Both of them are totally forgotten now, never having used them as neither is a practical subject.

Try to understand symmetry without group theory.

Ba'al Chatzaf

[Philip Coates]

Two others who have posted answers on this, Baal and David McK:

Did either of you also have a career in a math or engineering related field?

[DavidMcK]

No, I took calculus in college, failed it and had to take it over in the summer and finally made a 'C' (with an easy teacher). The feeling when I walked in and took the final the first time is hard to describe (dismayed maybe?). Your question was easy compared to that final.

[merjet]

DavidMcK, please tell us how you got the correct answer.

One doesn't really need calculus to get the correct answer. It could be done with geometry and pattern recognition.

The first set of numbers below shows that, for a given volume and height, the base being a square minimizes the surface.

The second set of numbers below shows that, for a given volume and the base a square, an 8x8 square minimizes the surface.

width___length___height__volume__surface (no top)

8.00____8.00____4.00____256____192.00

7.00____9.14____4.00____256____193.14

6.00____10.67___4.00____256____197.33

5.00____12.80___4.00____256____206.40

5.00____5.00___10.24____256____229.80

6.00____6.00____7.11____256____206.67

7.00____7.00____5.22____256____195.29

8.00____8.00____4.00____256____192.00

9.00____9.00____3.16____256____194.78

10.00___10.00___2.56____256____202.40

11.00___11.00___2.12____256____214.09

Edited August 22, 2010 by Merlin Jetton

[DavidMcK]

Actually I cheated and found a problem so similar online that I had the answer. Also I found this, a much more difficult way to solve it from someone with a screen name of HallsofIvy (an ivy leaguer maybe?):

Mar2-09, 01:56 PM

You can also use the "Lagrange multiplier" method. In order to minimize f(x,y,z), while subject to the condition g(x,y,z)= constant, then the gradients must be in the same direction- \nabla f[\itex] must be a multiple of [itex]\nabla g or \nabla f= \lambda\nabla g (\lambda is the "Lagrange multiplier")- so you differentiate both functions.

Here, f(x,y,z)= 2xy+ 2yz+ 2xz and g(x,y,z)= xyz= 256. \nabla f= (2y+ 2z)\vec{i}+ (2x+2z)\vec{j}+ (2y+ 2x)\vec{k} and \nabla g= yz\vec{i}+ xz\vec{j}+ xy\vec{k} so 2y+ 2z= \lambda yz, 2x+ 2z= \lambda xz, 2y+ 2x= \lambda xy. Since we don't really need to determine \lambda one method I like to solve equations like these is to divide one equation by another:

\frac{2y+ 2z}{2x+ 2z}= \frac{y}{x}

and

\frac{2x+ 2y}{2y+ 2x}= \frac{x}{z}

Those can be written as y^2+ zy= xy+ yz or y= x and xz+ yz= xy+ x^2 or z= x. That is, x= y= z (which is reasonable from symmetry considerations). Putting x= y= z into xyz= 256, we have x3= 256= 28= 43(4) so x= y= z= 4\sqrt[4]{4}. Edit: I found this criticism a few posts down from this: Mar2-09, 04:11 PM

HallsOfIvy: You're wrong, just because you didn't read the question carefully enough. It's a box with no top; your f(x) is wrong.

A few posts down from this complicated answer is a very simple answer (with no math) from csprof2000 (I found this today btw):

Mar2-09, 04:14 PM

Derill03:

I actually worked this out without doing a line of math, and I think you hearing it is good for you.

You probably already know that a cube has the lowest A/V ratio of any parallelpiped (box).

You're looking for what is essentially a parallelpiped sliced in half which, were it whole, would have area 512. The best one for you is an 8x8x8 cube.

Cut it in half, and you have an 8x8x4 shape with exactly the characteristics you wanted. If the cube was the best parallelpiped, the half-cube will be the best half-parallelpiped.

This is from here: http://www.physicsforums.com/archive/index.php/t-296518.html

Edited August 26, 2010 by DavidMcK