2013 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
let the length of side of base be y.

y = a - 2\frac{x}{tan{\frac{\pi}{6}}}

= a - \frac{2}{\frac{1}{\sqrt{3}}}

= a - 2x\sqrt{3}

Required Volume, V = \frac{1}{2} (a - 2x \sqrt{3})^{2} \text{sin}(\frac{\pi}{3}) x

= \frac{x\sqrt{3}}{4}(a - 2x \sqrt{3})^{2}

(ii)
\frac{dV}{dx} = \frac{\sqrt{3}}{4} (a - 2x \sqrt{3})^{2} -3x (a - 2x \sqrt{3})

For maximum volume, \frac{dV}{dx} = 0

\sqrt{3} (a - 2x \sqrt{3})^{2} -12x (a - 2x \sqrt{3}) = 0

(a - 2x \sqrt{3}) [\sqrt{3} (a - 2x \sqrt{3}) -12x] = 0

(a - 2x \sqrt{3}) = 0 ~\mathrm{or}~ a - 2x \sqrt{3}- 4x \sqrt{3} = 0

x = \frac{a}{2 \sqrt{3}}) ~\mathrm{or}~ a - 6x\sqrt{3} = 0

x = \frac{a}{2 \sqrt{3}} ~\mathrm{or}~ x = \frac{a}{6\sqrt{3}}

\frac{d^{2}V}{dx^{2}} = -3 (a - 2x \sqrt{3}) - 3 (a - 2x \sqrt{3}) + 6 \sqrt{3}x

\frac{d^{2}V}{dx^{2}} = -6a + 12x \sqrt{3} + 6\sqrt{3}x = -6a + 18x \sqrt{3}

When, x = \frac{a}{2 \sqrt{3}},~~ \frac{d^{2}V}{dx^{2}} = -6a + 18(\frac{a}{2 \sqrt{3}})\sqrt{3} = 3a > 0

When, x = \frac{a}{6 \sqrt{3}},~~ \frac{d^{2}V}{dx^{2}} = -6a + 18(\frac{a}{6 \sqrt{3}})\sqrt{3} = -3a < 0 Thus, Volume is maximum when x = \frac{a}{6 \sqrt{3}} \therefore, \mathrm{maximum~volume} = \frac{a^{3}}{54}

KS Comments:

Students MUST remember to the second order derivative, in order to which is the correct answer to accept.

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