4MA0 Higher January 2016 Paper 3R Q15
15 The diagram shows a cylinder inside a cone on a horizontal base.
The cone and the cylinder have the same vertical axis.
The base of the cylinder lies on the base of the cone.
The circumference of the top face of the cylinder touches the curved surface of the cone.
Diagram NOT accurately drawn
The height of the cone is 12 cm and the radius of the base of the cone is 4 cm.
Give your answer correct to 3 significant figures. (3)
The cylinder has radius \(r\) cm and volume \(V\) cm3
Find the value of \(r\) for which \(V\) is a maximum. (4)
| Scheme | Marks |
|---|---|
| \(l^2 = 12^2 + 4^2\) | M1 |
\(\pi \times 4 \times \sqrt{12^2 + 4^2}\) or \(\pi \times 4 \times \sqrt{160}\) or \(\pi \times 4 \times 12.6(4911064\ldots)\) or \(50.56\pi\) or \(\dfrac{1264}{25}\pi\) | M1 |
| 159 | A1 |
| (3) |
Notes
M1: For \(12^2 + 4^2\)
M1: For a correct expression of the curved surface area.
A1: Accept 158 – 159
| Scheme | Marks |
|---|---|
\(\dfrac{12 - h}{r} = \dfrac{12}{4}\) or \(4(12 - h) = 12r\) or \(\dfrac{h}{12} = \dfrac{4 - r}{4}\) or \(4 : 12 = r : 12 - h\) oe | M1 |
| \(h = \dfrac{12(4 - r)}{4}\) or \(h = 3(4 - r)\) or \(h = 12 - 3r\) | M1 |
| \(V = 12\pi r^2 - 3\pi r^3\) | A1 |
| (3) |
Notes
M1: For \(\dfrac{h}{12} = \dfrac{r}{4}\) or \(4h = 12r\) or \(4 : 12 = r : h\) oe
M1: \(h = 3r\)
A1: Dep. For \((V =)\;\pi r^2(12 - 3r)\)
Dep on M2
cso
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}r} = 24\pi r - 9\pi r^2\) | B2 |
| \(24\pi r - 9\pi r^2 = 0\) | M1 |
| \(2\tfrac{2}{3}\) | A1 |
| (4) |
Notes
B2: B1 for each term.
M1: ft their \(\dfrac{\mathrm{d}V}{\mathrm{d}r} = 0\)
A1: Accept \(\dfrac{24}{9}\), \(\dfrac{8}{3}\) oe