4MA0 Higher January 2014 Paper 4 Q15
15 A farmer has 120 metres of fencing.
He is going to make a rectangular enclosure \(PQRS\) with the fencing.
He is also going to divide the enclosure into two equal parts by fencing along \(MN\).
The width of the enclosure is \(x\) metres.
The length of the enclosure is \(y\) metres.
(a)
(i) Show that \(y = 60 - 1.5x\)
The area of the enclosure \(PQRS\) is \(A\) m2
(ii) Show that \(A = 60x - 1.5x^2\)
(3)(b) Find \(\dfrac{\mathrm{d}A}{\mathrm{d}x}\) (2)
(c) Find the maximum value of \(A\). (3)
| Scheme | Marks |
|---|---|
| (i) \(3x + 2y = 120\) | B1 |
| \(2y = 120 - 3x\) or \(1.5x + y = 60\) Answer: \(y = 0.5(120 - 3x)\) * | B1 |
| (ii) \(A = x \times y\) \(A = x \times (60 - 1.5x)\) Answer: \(A = 60x - 1.5x^2\) * | B1 |
| (3) |
Notes
B1: B1 dependent on first B1
* Answer given on question paper.
B1: * Answer given on question paper.
The mark scheme states that for this part the correct answer alone does not imply a correct method.
| Scheme | Marks |
|---|---|
| \((\mathrm{d}A/\mathrm{d}x =)\; 60 - 3x\) | B2 |
| (2) |
Notes
B2: If not B2 then B1 for 60 or \(-3x\)
| Scheme | Marks |
|---|---|
| “\(60 - 3x\)” = 0 | M1ft |
| \(x = 20\): (\(y = 30\)) | A1 |
| \(A = 20 \times 30\) or \(60 \times 20 - 1.5 \times 20^2\) Answer: 600 | A1 |
| (3) | |
| (8 marks) |
Notes
A1: cao Answer only = M1A2 (full marks)