# 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.

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Introduction

Maths Coursework 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. By Terry Whitcomb In this coursework I will be investigating whether the shopkeepers claim is correct. I want to find out if the volume of the tray will be a maximum if the area of the base is the same as the area of the four sides. To investigate their claim I will use tables to show my results. I will be investigating further by seeing if the shopkeeper's claim does work and if it does use it on other shapes such as rectangular trays. Firstly I am going to look at the example square and see if the shopkeeper's theory is correct: Side 18cm 18cm Lengths Volume 16x1x16 256 14x2x14 392 12x3x12 432 10x4x10 400 8x5x8 320 6x6x6 216 4x5x4 112 2x6x2 24 It is apparent that the 12x3x12 net creates the largest volume. Now lets see if the shopkeeper's theory is correct. Area of base-12x12=144 Area of sides-12x3x4=144 This shows that in this case the shopkeeper's claim is correct. There is not enough information yet to conclude from these results. ...read more.

Middle

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 18x3x18 972 16x4x16 1024 14x5x14 980 12x6x12 864 10x7x10 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. 12cmby12cm Square: Lengths Volume 10x1x10 100 8x2x8 128 6x3x6 108 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. ...read more.

Conclusion

� x (L�6) MV= maximum volume L=Length This formula gives me the maximum volume for any square: (L-2d) �xd 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 10x1x4=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 14cm 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 10x2x4=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. ...read more.

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