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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. In terms of L, x is 1/6 of L. When: x1/6L V= 1/6L(L-2/6L)² V= 1/6L(2/3L)² V= 1/6L X 2/3L X 1/6L V= 2/27L³ I will now substitute L=18cm into that equation: 2/6L=2/27 X 18³ =...

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