# 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.

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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

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

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