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RE: Optimization Project
Posted:
Dec 8, 1999 3:28 PM


The cost function is indeed c = .01Ph + .05A. The first step toward minimizing c is to make P as small as possible, if this can be done without affecting the base area or height (whose product gives the required volume.) This minimizing of P for given A is achieved by using a *regular* hexagon. For *any* given base area and height, using a nonregular hexagon will simply increase the cost of the lateral faces without any saving elsewhere and without any increase in volume.
In addition to the above equation, if we use x cm for a side of the hexagon, we have
P = 6x and A = 1.5 x^2 sqr(3). Using Ah = 355, as mentioned, and writing everything in terms of x gives:
c = 14.2/(x sqr(3)) + .075 sqr(3) x^2
For positive x, this function has one relative min, which is also the absolute min, found by setting dc/dx equal to zero:
(dc/dx =)  14.2/(x^2 sqr(3)) + .15 sqr(3) x = 0
x = cuberoot(14.2 / 0.45) = 3.16 approx, for which the cost is 3.89 cents approx.
I hope this is right <smile>
Greg
Original Message From: ownerapcalc@ets.org [mailto://ownerapcalc@ets.org] On Behalf Of Lynn A. Fisher Sent: Wednesday, December 08, 1999 9:50 AM To: apcalc@ets.org Subject: Re: Optimization Project
In regards to Greg's answer, I don't think it's that simple because the problem isn't simply to minimize area, it's to minimize cost. In fact, since the cost of the lateral area is less, it's reasonable to assume a contrary strategy...lower the area of the base and use a larger perimeter.
Here's where I got A=area of base P=perimeter of base h=height of solid
The cost function c(A,P,h) = .01Ph+.05A
Since h = 355/A
C(A,P) = 3.55P/A + .05A
And I see no relationship between A and P to put this in terms of a single variable.
I'm rusty on multivariable calculus, but I don't think there's a minimum for the cost function except if A=0. Anybody better than me at multivariable that wants to take it from there?
(In sum, I think the problem needs to have specified a regular base in order to have a solution.)
Lynn Fisher Woodstock Union HS Woodstock VT
Greg Spanier wrote:
> The surface area of the lateral faces depends upon the perimeter of the base > (and, of course, the height.) To minimize the perimeter of a polygon of any > given area, and hence the lateral area, the polygon needs to be regular. > However, I do think the question probably should have specified that the > hexagon be regular. > > Hope this helps, > > Greg > > Original Message > From: ownerapcalc@ets.org [mailto://ownerapcalc@ets.org] On Behalf Of > Mygirls810@aol.com > Sent: Monday, December 06, 1999 10:50 PM > To: apcalc@ets.org > Subject: Optimization Project > > Hi All, > > I recently assigned a project to my AB class and we are encountering some > confusion. If anyone could clarify things it would be appreciated. I got > the project from "A Watched Cup Never Cools" put out by Key Curriculum > Press. > The name of the project is PRISM POP. The set up is as follows: > > Your team has been given the assignment of submitting a packaging plan > for a new product. Prism Pop is a soda to be sold in hexagonally based > cans, > each holding 355 milliliters of pop.The management prefers plans that lower > the cost. The material for the sides costs 0.01 cents per square > centimeter. > The material for the bottom costs 0.03 cents per square centimeter. The > material for the top costs 0.02 cents per square centimeter. > > My question is this: the problem did not specify that the base had to be a > REGULAR hexagon even though the diagram accompanying it did show a regular > hexagon. If the base is not regular, how could you find the cost of the > area > to minimize? I may be missing something obvious, but I really am stumped on > this one. All help is greatly appreciated. > > Thanks, > Jeanne M. Benecke > Mygirls810@aol.com > Tappan Zee High School > Orangeburg, NY 10965 > (914)6801601



