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• Level: GCSE
• Subject: Maths
• Word count: 1675

Extracts from this document...

Introduction

Identical good tomatoes are placed in a box (see example).

Each tomato is a sphere. Each tomato just touches all the other tomatoes next to it as shown in the diagram. Tomato five goes bad. This is counted as the first hour. One hour later, all the tomatoes it touches go bad (now tomatoes 5,1,6 and 9 are bad). This continues every hour untill all the tomatoes in the box are bad.

I aim to investigate how tomatoes go bad in the above tray, and in trays of different sizes.

## How do tomatoes go bad in trays?

In this investigation, I aim to find a formula for calculating the total time required for all tomatoes to go bad in a rectangular tray of any rectangular size, and with any bad tomato starting position. To achieve this, I will first need to explain with the use of diagrams, how I arrived at a formula for the total time required for the tray to go bad. This will be done in three stages, followed by a worked example, which will answer both part 1 of the investigation, and part 2. Part 3 is an extension of the investigation concerned with the

Middle

Therefore the total time required to reach the corner is

If this were the most remote corner from the starting position, it would represent the total time needed for the whole tray to go bad.

Stage three of the analysis

Now we can calculate the time required for a rectangular tray of size M x N to go bad. A shaded cell represents the initial bad tomato.

Applying what we have learned in the previous stage of the analysis, we can firstly calculate the time required for the tomatoes in all four corners going bad, and secondly picking the longest time from these four results. The longest time, therefore will be the total time required for the whole tray to go bad. The formulas for calculating the time the bad tomatoes reach the corners are:

Corner T1 = (Q-1) + (P-1)

Corner T2 = (M-Q) + (P-1)

Corner T3 = (N-P) + (M-Q)

Corner T4 = (Q-1) + (N-P)

So the time needed for the whole tray to go rotten is

Worked example

This worked example will answer part 1 of the given investigation. The given tray size 4 by 4 can be seen in the following diagram, where the differently shaded cells are the three distinctively different starting positions.

Conclusion

T7=6

Tomato 11 is the initial bad tomato:

T11=6

Tomato 12 is the initial bad tomato:

T12=5

Tomato 13 is the initial bad tomato:

T13=4

So the average time for the tray to go bad is:

Taverage = [(8+7+6+6+5+4)/6] hours=6 hours

This can be seen easier on the following histogram:

Applying my formula for calculating the approximate average time for the tray to go bad:

Trough average = ¾(M+N) = ¾(5+5) = 7.5 hours

Again the result is that the approximate average is close to the true average different by 20%.

I predict that as the size of the tray increases, the true average time required for the tray to go bad and the approximate average time given by my formula will become closer.

Conclusion for whole investigation

I have develpoed a general formula for calculating the time for a tray of any size to become rotten, starting from one tomato. I applied the formula to the practical case of square size 4 by 4. The formula does not cover the case of several tomatoes going bad at the beginning in different places of the tray. I can only say that the tray will become rotten faster in this case. I have also devised a formula which calculates the approximate average time required for a tray of any size to go bad and tested it on trays 4 by 4 and 5 by 5 which show that it is accurate within  20%.

This student written piece of work is one of many that can be found in our GCSE Bad Tomatoes section.

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