2019 H2 Maths Paper 2 Q3 — Worked Solution

Solution

7 marks

Let $r$ cm be the radius and $h$ cm the height of the cylinder.

Approach

the volume depends on two variables linked by the fixed total surface area, so we use the surface area condition to write $h$ in terms of $r$ , substitute into $V$ to get a function of $r$ alone, and maximise using calculus.

The total surface area consists of the curved surface and the two circular ends:

2\pi r^{2} + 2\pi rh = 900

Making $h$ the subject,

h = \frac{900 - 2\pi r^{2}}{2\pi r} = \frac{450}{\pi r} - r

Substituting into $V = \pi r^{2}h$ ,

V

= \pi r^{2}\left( \frac{450}{\pi r} - r \right)

= 450r - \pi r^{3}

Differentiating with respect to $r$ ,

\frac{dV}{dr} = 450 - 3\pi r^{2}

At a stationary point, $\frac{dV}{dr} = 0$ :

450 - 3\pi r^{2} = 0

\Rightarrow r^{2} = \frac{150}{\pi}

\Rightarrow r = \sqrt{\frac{150}{\pi}}\ \ \left( r > 0,\ \text{rejecting the negative root} \right)

Since $\frac{d^{2}V}{dr^{2}} = - 6\pi r < 0$ for all $r > 0$ , this stationary point is a maximum.

Using $\pi r^{2} = 150$ ,

V_{max}

= 450r - \pi r^{3}

= r\left( 450 - \pi r^{2} \right)

= r(450 - 150)

= 300r

Since $r = \sqrt{\frac{150}{\pi}} = \frac{5\sqrt{6}}{\sqrt{\pi}} = \frac{5\sqrt{6\pi}}{\pi}$ ,

V_{max} = \frac{1500\sqrt{6\pi}}{\pi}

For the ratio $r:h$ , using $\frac{450}{\pi} = 3\left( \frac{150}{\pi} \right) = 3r^{2}$ (since $r^{2} = \frac{150}{\pi}$ ),

h = \frac{450}{\pi r} - r = \frac{3r^{2}}{r} - r = 3r - r = 2r

To check your answer:

2\pi r^{2} + 2\pi rh = 2\pi r^{2} + 2\pi r(2r) = 6\pi r^{2} = 6(150) = 900

, matching the given surface area. ✓

$\therefore$ the maximum possible volume is $V = \frac{1500\sqrt{6\pi}}{\pi}\ cm^{3}$ (equivalently $1500\sqrt{\frac{6}{\pi}}\ cm^{3}$ , $\text{≈}2070\ cm^{3}$ ), attained when $r:h = 1:2$ .

2019 H2 Maths Paper 2, Question 3

Solution

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