The shop keeper says, "When the area of the base is the same as the area of the four sides, the volume of the tray will be a maximum"Investigate this claim.

From the table it is evident that the maximum volume available will be from a net 24x6x24. I am now going to see if this tray follows the shopkeeper's claim as well.

Area of base-24x24=576

Area of sides-24x6x4=576

This square also follows the shopkeeper's theory.

I am going to list the similarities the previous 2 squares had to help me in my investigation:

Both in the 2X table

Both in the 3X table

Both in the 6X table

Both even numbers

36 is double 18

I am now going to investigate a 9cmx9cm square and see if it follows the shopkeeper's claim, as it is half of 18.

9cmx9cm Square:

Lengths

Volume

7x1x7

49

5x2x5

50

3x3x3

27

x4x1

4

The table shows that you would get the maximum volume from the net 5x2x5.

I am now going to work out whether this square follows the shopkeeper's rule.

Area of base-5x5=25

Area of sides-5x2x4=40

This square does not follow the shopkeeper's claim.

This square is different to the 18cmx18cm and the 36cmx36cm squares because:

* The lengths are odd numbers

* The lengths are not in the 6X table

Maybe this square does not apply to the shopkeeper's theory because it is a odd number or it is not part of the 6X table.

I am going to investigate a even number to see if the shopkeeper's claim applies to that square.

I am going to try a 28cmx28cm square:

Lengths

Volume

26x1x26

676

24x2x24

152

22x3x22

452

20x4x20

600

8x5x18

620

6x6x16

536

4x7x14

372

2x8x12

152

The table indicates that the maximum volume would be give with the net 18x5x18.

Does the square follow the shopkeeper's claim?

Area of base-18x18=324

Area of sides-18x5x4=360

This square does not follow the shopkeeper's claim. This square had even lengths.

This leaves me to believe that the shopkeeper's claim only works with squares that have lengths that are multiples of 6.

I am going to do some more squares that are multiples of 6 to confirm this theory.

24cmx24cm Square:

Lengths

Volume

22x1x22

484

20x2x20

800

8x3x18

972

6x4x16

024

4x5x14

980

2x6x12

864

0x7x10

700

The maximum volume would be given from the net 16x4x16.

Does this square follow the shopkeeper's claim?

Area of base-16x16=256

Area of sides-16x4x4=256

This square supports the shopkeeper's claim. This square is a multiple of 6. It seems that the shopkeeper's claim only applies to factors of 6. I will investigate my theory by looking at more multiples of 6.

2cmby12cm Square:

Lengths

Volume

0x1x10

00

8x2x8

28

6x3x6

08

The maximum volume would be given from the net 8x2x8.

I suspect that this square will follow the shopkeeper's claim.

Area of square-8x8=64

Area of sides-8x2x8=64

This square also supports the shopkeeper's claim. To confirm that the shopkeeper's claim does work on all multiples of 6 I am going to investigate one more square.

60cmx60cm:

Lengths

Volume

58x1x58

3364

56x2x56

6272

54x3x54

8748

52x4x52

0816

50x5x50

2500

48x6x48

3824

46x7x46

4812

44x8x44

5488

42x9x42

5876

40x10x40

6000

38x11x38

5884

36x12x36

5552

34x13x34

5028

32x14x32

4336

The maximum volume would be given from the net 40x10x40.

To confirm my theory I will see if this square follows the shopkeeper's claim

Area of base-40x40=1600

Area of sides-40x10x4=1600

The square 60cmx60cm also backs up the shopkeeper's claim. I am going to look at the results I have collected from the multiples of 6 to see if I can find any relationships:

Length of side:

8cmx18cm

36cmx36cm

2cmx12cm

24x24cm

60cmx60cm

Net that gives highest volume

2x3x12

24x6x24

8x2x8

6x4x16

40x10x40

Relationships found:

The net that has the highest volume has always got the depth 4 times larger than the length. Example -12x3x12- 3 goes into 12, 4 times.

The net that has the maximum volume has a length that is always two thirds of the original length. Example-original length 36cm, Maximum volume net- 24cm. 12 goes in 36 3 times, 12 goes into 24 2 times.

The depth is always how many times 6 goes into the original length. Example-12cmx12cm. 6 goes into 12 2 times so the depth of the tray that will have maximum volume will be 2cm.

I can now tell instantly what the maximum volume will be from any square with the lengths that are multiple of 6.

For example:

48cmx48cm:

Two thirds of 48 is 32, I now the length will be 32cm.

Now I can either divide 32 by 4, which will give me 8 or see how many times 6 goes into 48 which is 8. This gives me the depth.

I now know that the net that gives me the maximum volume will be 32x8x32.

This gives me the maximum volume of the tray, which is 8192.

I can now create a formula to give the maximum of any square tray that is a multiple of 6:

MV=(Lx2÷3) ² x (L÷6)

MV= maximum volume

L=Length

This formula gives me the maximum volume for any square:

(L-2d) ²xd

L-Length

D-Depth

My next investigation is to see whether this claim works on rectangles.

I have chosen a 6x12 rectangle as both lengths are in the 6X table and will show me if the theory works on rectangles I am unable to use the formula for the squares on the rectangles because there are two lengths.

6cm sides

12cm

Lengths

Volume

4x1x10

40

2x2x8

32

It is apparent that the net 4x1x10 gives the maximum volume. I am going to see if the shopkeeper's theory works on this rectangle and see if the formula works on it.

4x10=40

0x1x4=40

The shopkeeper's theory does work on this rectangle. I am now going to investigate more rectangles to see if I can find any patterns.

I am now going to add 2cm to each length and see the result.

8cm

Sides

4cm

Lengths

Volume

6x1x12

72

4x2x10

80

2x3x8

48

4x2x10 appears to give the largest volume. I am going to see if the shopkeeper's claim is correct in this case.

4x10=40

0x2x4=80

It appears that the shopkeeper's claim does not work on this rectangle; maybe it does not work because neither number is in the 6x table.

I am going to record my results of various rectangles to see if I find any patterns.