Higher November 2023 Paper 3 Q23
23 A chocolate box in the shape of a prism is being designed.
All lengths are in centimetres.
The cross section is a regular hexagon with side \(x\)
The length is \(5x\)
An expression for the area of the cross section, in cm\(^2\), is \(\;\dfrac{3\sqrt{3}}{2}x^2\)
The total surface area of the box must be less than 650 cm\(^2\)
Work out the largest possible integer value of \(x\).
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x \times 5x\) or \(5x^2\) | M1 | oe may be implied eg \(30x^2\) |
| \(2 \times \dfrac{3\sqrt{3}}{2}x^2 + 6 \times x \times 5x\) or \(3\sqrt{3}\,x^2 + 30x^2\) | M1dep | oe eg \(35.19(6…)x^2\) or \(35.2x^2\) |
| \(650 \div (3\sqrt{3} + 30)\) or [18.4, 18.5] or [4.2, 4.3] or \(3\sqrt{3} \times 4^2 + 30 \times 4^2\) or 563.(…) and \(3\sqrt{3} \times 5^2 + 30 \times 5^2\) or 879.(…) or 880 | M1dep | oe dep on M2 calculation or [18.4, 18.5] may be seen in a square root trials \(x = 4\) and \(x = 5\) ignore substitution of other integer values of \(x\) |
| 4 with at least first two M marks awarded | A1 |