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• Level: GCSE
• Subject: Maths
• Word count: 1004

The shopkeeper says, 'when the area of the base is the same as the area of the four sides the volume will be at a maximum'. Investigate the shopkeepers claim and see if it is correct.

Extracts from this document...

Introduction

Trays

A shopkeeper asks a company to make some trays. The shopkeeper says, ‘when the area of the base is the same as the area of the four sides the volume will be at a maximum’.

I will use different mathematical means to investigate the shopkeepers claim and see if it is correct.

Formula

V= Volume                        HLW

W= Area of walls        4(L X H)

B= Area of base                LW

Investigation 1

 Height Length Width Walls Base Volume 1 16 16 64cm² 256cm² 256cm² 2 14 14 112cm² 196cm² 392cm² 3 12 12 144cm² 144cm² 432cm² 4 10 10 160cm² 100cm² 400cm² 5 8 8 160cm² 64cm² 216cm² 6 6 6 144cm² 36cm² 112cm² 7 4 4 112cm² 16cm² 32cm² 8 2 2 64cm² 4cm² 0cm² 9 0 0 0cm² 0cm² 0cm²

As you can see the above shows that when the height of the sides equals 3. the trays volume is at a maximum. This shows the shopkeepers claim to be correct so far as the area of all four sides equals the area of the base.

I will now use a formula to further test the shopkeeper’s theory:

(L-2x)²        =        4x(L-2x)

÷(L-2x)                ÷(L-2x)

L-2x                =        4x

L                =        6x

X                =        1/6L

This shows the height in terms of length.

Middle

Length

Width

Walls

Base

Volume

2.9

12.2

12.2

141.52cm²

148.84cm²

431.636cm³

3.1

11.8

11.8

146.32cm²

139.24cm²

431.644cm³

2.95

12.1

12.1

142.78cm²

146.41cm²

431.9095cm³

3.15

11.7

11.7

147.42cm²

136.89cm²

431.2035cm³

Results in the table so far agree that 432cm³ is the maximum volume.

Investigation 2

I will now use a tray using Length=15cm.

This can also be used to prove the shopkeeper is correct.

(15-2x)²        =        4(15-2x)

÷(15-2x)                ÷(L-2x)

15-2x        =        4x

15                =        6x

X                =        2.5

V                =        (15-2x)(15-2x)

V                =        (225-30x+4x²)x

V                =        225x-60x²+4x³

V                =        250cm³

This formula shows that when the height of the sides equal 2.5, the trays volume will be at its maximum.

Investigation 3

To further my investigation on the 18x18cm tray I will use differentiation. Differentiation finds the maximum point. By differentiating a function we get the value of any gradient at any particular point. This is called the derivative.

Y                =        Volume

x                =        Height

x                =        18

y                =        x(18-2x)²

=        x(324-72x+4x²)

=        324-72²+4x³

Dy/dx        =        324-144x+12x²

Maximum        =        324-144x+12x²=0

(18-2x)²-72x+12x²=0

Area of base=        (18-2x)² = 72x-8x²

Area of side=        4x(18-2x)

Conclusion

(18-2x_(12-2x)                =        2x(18-2x)+2x(12-2x)

216-60x+4x²                =        36x-4x²+24x-4x²

12x²-120x+216                =        0

X²-10x+18                        =        0

A=1

B=-10

C=18

X=

Not possible                        Where decimal search would end up

Base area=

Area of sides=

Volume=

After investigation number one I found that the shopkeepers claim to be correct using a square based try and substituting lengths into a formula. I further checked the shopkeepers claim by using a trial and error table. This also showed the theory to be correct.

I also tried the same investigation method in investigation two but this time using a 15x15 square. This also showed the claim to be correct.

I then tried differentiation on a 18x18 sqaure. This showed the volume (y) is at a maximum when the area of the base and the area of the sides are equal. Therefore confirming the claim to be correct.

I conclude that the shopkeepers claim is correct for both square and rectangular trays after carrying out my investigations using varying mathematical means.

Robert Davison

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