Chief Delphi - Y=ax^2+bx+c Fact or Fiction?

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Bduggan04

05-01-2003 22:15

Quote:

Originally posted by Johca_Gaorl
Actually correction, if it's odd, it doesn't matter whether more are in the stack or on the ground:

i.e. 21 total

11 stack * 10 ground = 110
10 stack * 11 ground = 110

This was mentioned before. I think they said more on the ground because it was easier to achieve and a shorter stack is more stable.

Mark Garver

05-01-2003 22:56

Actually....

If you have an odd number of boxes this isn't always the case because of the orientation in which the boxes could be placed. Meaning you stack the orientation of the boxes to your advantage in stacking the boxes. Maximizing the number scored. (ie. the highest score is actually the maximum number of boxes possible (not 45) and dividing by 2).

Johca_Gaorl

06-01-2003 06:39

Re: Actually....

Quote:

Originally posted by Mark G
If you have an odd number of boxes this isn't always the case because of the orientation in which the boxes could be placed.

Right, I forgot to include that part.

junkyarddawg

06-01-2003 14:11

1 Attachment(s)

You can tabulate and graph this real easy in Excel. Make three columns. Column B is the tubs in the highest stack, column C is the number of tubs not in the multiplier stack and column D is the total score. Put 45 at the top of colmn B and 0 at the top of column C and put the formula =B*C at the top of column D.
In second cell of column B put B1-1 and in Column C put C1+1 and repeat the formula from D1 to D2. Copy B1 through D1 and paste it in the next 44 cells below B2.

You should end up with a column of descending numbers 45 - 0 and a column of ascending numbers 0 - 45 and the third column should be the product.

And, as you'll see, you'll want your tallest stack to equal half or in the case of an odd number of tubs, one less than half to get the highest score. Not forgetting that the top of the stack is a tub on end or upside down. I attached an example. Play with it.

maDGag

09-01-2003 22:51

discreet

tisk tisk tisk

can't do that ... no maximization

its discrete

but ... u can base ur answer the fact that the maximum area of a paralellogram is in a square ... hence the (x/2)^2 yielding max points

Jay Lundy

10-01-2003 03:10

Quote:

Originally posted by Suneet
I already posted this in a different thread, but...

y=a*x^2+b*x+c

y= raw score
a=-1
x=Number of boxes in your highest stack
b=Total number of boxes on your side
c=0, 25, or 50, depending on the #of bots in the middle for you.

I'll admit it's not quite accurate, but that's not my fault.

Yes that's correct. Remember that when a parabola opens down (as in this case) the vertex is the maximum y value (y = total points).

The equation for the x coordinate of the vertex of any parabola is x = -b / (2 * a)

a = -1
x = -b / -2
x = b / 2
so....
Number of boxes in your highest stack = Total number of boxes on your side / 2

No calculus, just Algebra II.

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