# Write-up 12

### by

## Lynne Bombard

**Problem: **Place four numbers in the first row as follows: A B C D. For each
successive row replace the entries by the absolute value of the difference
of the entry just above and the entry just to the right in the previous
row. In the fourth position use the absolute vale of the difference of the
fourth and the first (i.e. cycle). abs.A-B, abs B-C, abs C-D, and abs D-A.

A.) Will the process lead to a 0 in all 4 entries for some
row?

B.) What is the largest number of rows before a zero row
is generated?

**Hint**: If answer
is less than 10, you should try again.

To answer part a, the process will lead to a 0 in all four
entries. The only difference is where that row of zeros appear. What makes
the difference is that by taking the absolute value you will never have
negative numbers and there will always be a way to get a row of all zeros.
For example, let's start with 1, 2, 3, and 4. Then I followed the steps
as mentioned above.

**A** |
**B** |
**C** |
**D** |

1 |
2 |
3 |
4 |

1 |
1 |
1 |
3 |

0 |
0 |
2 |
2 |

0 |
2 |
0 |
2 |

2 |
2 |
2 |
2 |

0 |
0 |
0 |
0 |

As you can see, the sixth row is a row of zeros. As you
notice, we will get a row of zeros when all four entries become the same
number. This would be the only way to get a row of zeros.

After this I tried increasing the values by increments
of one, but did not get very far. I began trial and error to see what types
of conclusions I could come up with. I knew that I had to keep at it until
I at least came up with four numbers that gave me more than 10 nonzero rows.
Two examples that I came up with follow below.

**A** |
**B** |
**C** |
**D** |

66 |
88 |
100 |
25 |

22 |
12 |
75 |
41 |

10 |
63 |
34 |
19 |

53 |
29 |
15 |
9 |

24 |
14 |
6 |
44 |

10 |
8 |
38 |
20 |

2 |
30 |
18 |
10 |

28 |
12 |
8 |
8 |

16 |
4 |
0 |
20 |

12 |
4 |
20 |
4 |

8 |
16 |
16 |
8 |

8 |
0 |
8 |
0 |

8 |
8 |
8 |
8 |

0 |
0 |
0 |
0 |

and the following is a second example that I came up.

**A** |
**B** |
**C** |
**D** |

67 |
89 |
101 |
26 |

22 |
12 |
75 |
41 |

10 |
63 |
34 |
19 |

53 |
29 |
15 |
9 |

24 |
14 |
6 |
44 |

10 |
8 |
38 |
20 |

2 |
30 |
18 |
10 |

28 |
12 |
8 |
8 |

16 |
4 |
0 |
20 |

12 |
4 |
20 |
4 |

8 |
16 |
16 |
8 |

8 |
0 |
8 |
0 |

8 |
8 |
8 |
8 |

0 |
0 |
0 |
0 |

In these two example, you can see that I came up with four
numbers that give me thirteen nonzero rows and therefore the fourteenth
row is zeros.

**Problem** : Explore
problems of maximization such as the lidless box formed from a 5x8 sheet
with a square removed from each corner.

**Length of Cut** |
**Area of base** |
**Volume of Box** |

0.25 |
33.75 |
8.4375 |

0.5 |
28 |
14 |

0.75 |
22.75 |
17.0625 |

1 |
18 |
18 |

1.25 |
13.75 |
17.1875 |

1.5 |
10 |
15 |

1.75 |
6.75 |
11.8125 |

2 |
4 |
8 |

2.25 |
1.75 |
3.9375 |

You can see from the above table that the maximum volume
is when x=1 the volume is 18.

**Problem**: Find
the maximum for f(x) = (1-x)(1+x)^2 on the interval [0,1].

**x** |
**(1-x)** |
**(1+x)^2** |
**(1-x)(1+x)^2** |

0 |
1 |
1 |
1 |

0.25 |
0.75 |
1.5625 |
1.171875 |

0.5 |
0.5 |
2.25 |
1.125 |

0.75 |
0.25 |
3.0625 |
0.765625 |

1 |
0 |
4 |
0 |

From the above table so far we see that the maximum is
going to be somewhere between .25 and .50.

**x** |
**(1-x)** |
**(1+x)^2** |
**(1-x)(1+x)^2** |

0.333333333333333 |
0.666666666666667 |
1.77777777777778 |
1.18518518518519 |

By doing Calculus I knew that the answer was going to be 1/3.
When I tried that value on Excel you can see that 1/3 is the maximum of
f(x)=(1-x)(1+x)^2.
I could have done more values in my spreadsheet that were
between .25 and .50, but it is easy to see that the value (from Calculus)
was 1/3.

**Return**