• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28
  29. 29
    29
  30. 30
    30
  31. 31
    31
  32. 32
    32
  33. 33
    33
  34. 34
    34
  35. 35
    35
  36. 36
    36
  37. 37
    37
  38. 38
    38
  39. 39
    39
  • Level: GCSE
  • Subject: Maths
  • Word count: 10642

maxi products

Extracts from this document...

Introduction

GCSE Maths Coursework  - Maxi Product

Introduction

In this investigation, I am going investigate the Maxi Product of numbers. I am going to find the Maxi Product for selected numbers and then work out a general rule after individual rules are worked out for each step. I am going to find the Maxi Product for double numbers, I will find two numbers which added together equal the number selected and when multiplied will equal the highest number possible that can be retrieved from two numbers multiplied together. I am also going to find the Maxi Product for triple numbers, I will find three numbers which added together equal the number selected and when multiplied will equal the highest number possible that can be retrieved from three number multiplied together. And finally, I am going to find the Maxi Product for quadruplet numbers, I will find four numbers which added together equal the number selected and when multiplied will equal the highest number possible that can be retrieved from four numbers multiplied together. After working out the individual rules for these three sectors of numbers, I will then work out the general rule for any amount of numbers it can be split into. For example, it can be split up into five numbers and I will be able to find the Maxi Product of any number given by splitting it up into five numbers. I will be using whole numbers, decimal numbers and fractional numbers.

Double Numbers

Examples: 12

(5,7)= 12  5+7  5x7=35

(6,6)= 12  6+6  6x6=36

(4,8)= 12  4+8  4x8=32

I have found that 36 is the highest number so far that can be retrieved from 6 and 6 when the number is 12, in whole numbers. I will now try in decimal numbers if I can get a number higher than 36.

(6.5,5.5)=12  6.5+5.5  6.5x5.5=35.75

(6.7,5.3)=12  6.7+5.3  6.7x5.3=35.51

(6.3,5.7)=12  6.3+5.7  6.3x5.7=35.91

(6.2,5.8)=12  6.2+5.8  6.2x5.8=35.96

(6.1,5.9)=12  6.1+5.9  6.1x5.9=65.99

...read more.

Middle

 4.1+4.8+4.1  4.1x4.8x4.1=80.688

(4.2,4.6,4.2)= 13  4.2+4.6+4.2  4.2x4.6x4.2=81.144

(4.3,4.4,4.3)= 13  4.3+4.4+4.3  4.3x4.4x4.3=81.356

I will now move onto fractional numbers as there can be no other decimal number that can give a result higher than 81.356 using three numbers. I will see in fractional numbers if I can get a number higher than 81.356 from three fractional numbers.

(4 1/2 ,4 1/4,4 1/4)=       13  4 1/2+4 1/4+4 1/2        4 1/2x4/14x4 1/4        =81.28125

(4 5/16,4 5/16, 4 6/16)= 13  4 5/16+4 5/16+4 6/16  4 5/16x4 5/16x 4 6/16=81.364

                                                                                                                                   (3dp)

(4 1/3,4 1/3,4 1/3)=        13  4 1/3+4 1/3+4 1/3        4 1/3x4 1/3x4 1/3       =81.370    

                                                                                                                                   (3dp)

I have found that 4 1/3, 4 1/3, and 4 1/3 are the three numbers which added together make 13 and when multiplied together make 81.370 (3dp) which is the highest possible answer which can be retrieved when three numbers added together equal 13 are multiplied.

14

(1,12,1)= 14  1+12+1  1x12x1=12

(2,11,1)= 14  2+11+1  2x11x1=22

(2,10,2)= 14  2+10+2  2x10x2=40

(3,9,2)=   14  3+9+2    3x9x2  =54

(3,8,3)=   14  3+8+3    3x8x3  =72

(4,7,3)=   14  4+7+3    4x7x3  =84

(4,6,4)=   14  4+6+4    4x6x4  =96

(5,5,4)=   14  5+5+4    5x5x4  =100

I am now going to use decimal numbers as I have found the highest possible result in whole numbers. I will see in decimal numbers if I can retieve a higher result than 100 when three numbers are multiplied together.

(4.7,4.9,4.4)= 14  4.7+4.9+4.4  4.7x4.9x4.4=101.332

(4.7,4.7,4.6)= 14  4.7+4.7+4.6  4.7x4.7x4.6=101.614

I will now move on to fractional numbers as there can be no other decimal number that can give a result higher than 101.614 using three numbers. I will see in fractional numbers if I can get a number higher than 101.332 from three fractional numbers.

(5 1/3,4 1/3,4 1/3)= 14  5 1/3+4 1/3+4 1/3  5 1/3x4 1/3x4 1/3= 100.148 (3dp)

(4 2/3,4 2/3,4 2/3)= 14  4 2/3+4 2/3+4 2/3  4 2/3x4 2/3x4 2/3= 101.6296296

I have found that 4 2/3, 4 2/3, and 4 2/3 are the three numbers which added together make 14 and when multiplied together make 101.6296296 which is the highest possible answer which can be retrieved when three numbers added together equal 14 are multiplied.

15

(1,13,1)= 15  1+13+1  1x13x1=13

(2,12,1)= 15  2+12+1  2x12x1=24

(2,11,2)= 15  2+11+2  2x11x2=44

(3,10,2)= 15  3+10+2  3x10x2=60

(3,9,3)=   15  3+9+3    3x9x3  =81

(4,8,3)=   15  4+8+3    4x8x3  =96

(4,7,4)=   15  4+7+4    4x7x4  =112

(5,6,4)=   15  5+6+4    5x6x4  =120

(5,5,5)=   15  5+5+5    5x5x5  =125

I am now going to use decimal numbers as I have found the highest possible result in whole numbers. I will see in decimal numbers if I can retieve a higher result than 125 when three numbers are multiplied together.

(5.1,4.8,5.1)= 15  5.1+4.8+5.1  5.1x4.8x5.1=124.848

(5.3,4.4,5.3)= 15  5.3+4.4+5.3  5.3x4.4x5.3=123.596

...read more.

Conclusion

(1,1,1,5)= 8  1+1+1+5  1x1x1x5=5

(2,1,1,4)= 8  2+1+1+4  2x1x1x4=8        

(2,2,1,3)= 8  2+2+1+3  2x2x1x3=12

(2,2,2,2)= 8  2+2+2+2  2x2x2x2=16

I will now move on to decimals as I have found the highest result in whole numbers. I will see in decimal numbers if I can retriev a result higher than 16 when four decimal numbers are multiplied.

(2.1,2.1,2.1,1.7)= 8  2.1+2.1+2.1+1.7  2.1x2.1x2.1x1.7=15.7437

(2.2,2.2,2.2,1.4)= 8  2.2+2.2+2.2+1.4  2.2x2.2x2.2x1.4=14.9072

I will now move on to use fractions as I have yet not found a number higher than 16. I will see in fractional numbers if I can retriieve a higher number than 16 when four numbers are multiplied together.

(2 5/10,2 2/10,2 1/10,1 2/10)=              8  2 5/10+2 2/10+2 1/10+1 2/10  2 5/10x2     

                                                               2/10x2 1/10x1 2/10=13.86 

(2 40/100,2 30/100,2 22/100,1 8/100)= 8  2 40/100+2 30/100+2 22/100+1 8/100

                                                               2 40/100+2 30/100+2 22/100=13.234752

I have found that 2,2,2 and 2 are the four numbers, which added together make 8 and when multiplied together make 16 which is the highest possible answer which can be retrieved.

Rule in words:

Maxi Product equals a fourth of the selected number multiplied by itself four times.

Rule in Algebra:

M=(N/4)image00.jpg

Key:

M= Maxi Product

N=Number Selected

Proving my rule:

Find the Maxi Product of a)44 b)55 and c)66.

a)M=(N/4)image00.jpg                b) M=(N/4)image00.jpg                     c) M=(N/4)image00.jpg

M=(44/4)image00.jpg                   M=(55/4)image00.jpg                        M=(66/4)image00.jpg

M=14644                       M=35744.62891                M=74120.0625

The General Rule:

Rule for doubles        M=(N/2)²                

Rule for triples           M=(N/3)³   

Rule for Quadruplets M=(N/4)image00.jpg

I notice that what ever the number is divided by, it is always then powered by the same number.

For example, if the number were to be split into 6 parts (sextuplets), then the equation for that would be: M=(N/6)image05.jpg .  

Rule in words:    

Maxi Product equals the number divided by X number of parts and that powered by X.

Rule in Algebra:

M=(N/X)image01.jpg

Key:

M=Maxi Product
N=Number selected
X=Parts it has been split up in e.g. Doubles, triples, etc.
Proving My Rule:     
Find the Maxi Product of a) 44 when it’s split up into 7 parts; b) 55 when it’s split up by 8 parts and c) 66 when it’s split up into 9 parts.
a)M=(N/X)image01.jpg
       M=(44/7)image02.jpg
       M=387688.0863
b)M=(N/X)image01.jpg
M=(55/8)image03.jpg
       M=4990931.629
c)M=(N/X)image01.jpg
       M=(66/9)image04.jpg

              M=61335630.63

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Miscellaneous essays

  1. Marked by a teacher

    Frogs Investigation - look at your results and try to find a formula which ...

    hops to the right Step 14-A1 hops to the right Step 15 -B1 slides to the left Step 16 -B2 hops to the left Step 17 -B3 hops to the left Step 18 -B4 hops to the left Step 19 -A3 slides to the right Step 20-A2 hops to the

  2. Stair Totals coursework

    my number stairs, as you go across the number increases by 1 and as you go up the number increases by 10. I will illustrate this below. + 10 +10 + 1 + 1 I will now do the same thing using algebra +1 + 10 + 10 + 1 + 1 T= n+ (n+1)

  1. GCSE module 5 AQA Mathematics

    2 x 4 boxes. For the past two formulas I have used a similar method and they have both been correct so I predict that for 2 x 4 boxes the equation will be: y= n + 10(n+3) - x= n (n+13)

  2. Trays. The first square I will investigate is a 24cm x 24cm square. My ...

    above proves that the shopkeeper's statement is false for a 26cm x 26cm square. Now I have discovered that the shopkeeper's statement isn't true for all sized squares I will investigate which squares the shopkeepers statement is true for. I have noticed that the squares that I have already investigated

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work