Bad Tomatoes

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Natalie Hayes 10E group 3                                                                                    Page 1

Bad Tomatoes

I have been given a box filled with identical good tomatoes, one of the tomatoes goes bad and after 1 hour, all the tomatoes that it touches go bad.  After another hour, the bad tomatoes make all the good tomatoes they touch go bad.  This continues until all the tomatoes in the box go bad.  

My aim in this investigation is to find a formula to show how many tomatoes in a tray go bad.

I have started the investigation by drawing diagrams to help get me started, I thought it would be more useful if I investigated how many tomatoes would go bad if the first tomato to go bad was in a different position i.e. the corner.  For the first part of the investigation I decided to use a tray that had unlimited sides.

         Corner                            Side                          Middle            

As you can see from the diagrams that the amount of tomatoes that go bad depends on the position of the first bad tomato.  If I was going to investigate the amount of tomatoes that would go bad win a tray with unlimited sides then I would have to draw myself bigger diagrams to help me find a sequence (look at page 6)  

Corner

Here is a table to show the total amount of bad tomatoes after a certain amount of hours, with the bad tomato starting in the corner in a box with unlimited sides.

By looking at this table of results, I can now produce a sequence.

1              2              3              4               5    

3              6             10            15             21

      +3             +4           +5             +6

              +1             +1             +1

Natalie Hayes 10E group 3                                                                                    Page 2

The first thing I noticed when looking at my sequence was that the difference between each number increased by 1 each time e.g. +1, +2, +3, +4, +5, +6.

As you can see from the sequence, second differences are present.  I used the formula An² + Bn + C to help me work out an nth term.  The first thing I did was to work out the simultaneous equations as I knew that it would find what A, B an C represented.

  1.     A  +    B  +  C  =  3
  2.   4A  +  2B  +  C  =  6
  3.   9A  +  3B  +  C  =  10
  4. 16A  +  4B  +  C  =  15

the next stage was to take the equations away from each other so I was left with two equations that would cancel each other out so I was left with what A represented:

   2.        4A  +  2B  +  C  =  6

  -  1.          A  +    B  +  C  =  3

     5.        3A  +    B          =  15

    4.      16A  +  4B  +  C  =  15

   -  3.        9A  +  3B  +  C  =  10

Join now!

      6.        7A   +   B           =  5

I then needed to take equation 5. away from equation 6. so that B would be cancelled out and I would be left with A.

       6.        7A   +   B           =  5

    -  5.        3A  +    B           =  15

                   4A               ...

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