# Math

posted by George

A home owner goes to the hardware store to purchase the letters that make up a number for a sign he is making, (e.g., F-O-U-R, N-I-N-E, etc.). When he arrives, other customers are doing the same thing. The first customer buys the letters to display ONE, and pays \$40. The next customer buys the letters for the number TWO, and pays \$60. The last customer buys the letters for the number ELEVEN, and pays \$100. Our home owner wants to buy the letters to spell out TWELVE.

The question is: how much does the home owner pay?

1. Reiny

ok George, I'll play your game.

I will assume that the store owner is some math-nerd and sets the price of the numbers according to some cubic equation
cost = ax^2 + bx + c, where x is the number of letters and a,b,c are constants
so we have three ordered pairs:
(1,40), (2,60) and (11,100)
for (1,40) ---> a + b + c = 40
for (2,60) ---> 4a + 2b + c = 60
for (11,100) -> 121a + 11b + c = 100

2nd - 3rd: ---> 3a + b = 20
3rd - 1st: ---> 120a + 10b = 60 or 12a + b = 6
subtracting those last two:
9a = -14
a = -14/9
from 12a+b=6
b = 6-12a = 6-12(-14/9) = 74/3

back in 1st: a+b+c=40
-14/9 + 74/3 + c = 40
c = 152/9

so cost = (-14/9)x^2 + (74/3)x + 152/9

so when x = 12, Cost = \$88.89

2. Reiny

I meant to say:

I would need 4 ordered pairs for a cubic.
My work reflects what is needed for a quadratic

3. George

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