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Topic: Optimization Project
Replies: 10   Last Post: Dec 10, 1999 9:49 AM

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Greg Spanier

Posts: 66
Registered: 12/6/04
RE: Optimization Project
Posted: Dec 8, 1999 3:28 PM
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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 non-regular 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

I hope this is right <smile>


-----Original Message-----
From: [mailto://] On Behalf Of Lynn
A. Fisher
Sent: Wednesday, December 08, 1999 9:50 AM
Subject: Re: Optimization Project

In regards to Greg's answer, I don't think it's that simple because the
isn't simply to minimize area, it's to minimize cost. In fact, since the
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

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
wants to take it from there?

(In sum, I think the problem needs to have specified a regular base in order
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
> (and, of course, the height.) To minimize the perimeter of a polygon of
> 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: [mailto://] On Behalf Of
> Sent: Monday, December 06, 1999 10:50 PM
> To:
> 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

> 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

> this one. All help is greatly appreciated.
> Thanks,
> Jeanne M. Benecke
> Tappan Zee High School
> Orangeburg, NY 10965
> (914)680-1601

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