2010 A-level H2 Mathematics (9740) Paper 1 Question 9 Suggested Solutions

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

(i)
Volume, V $= 3xxy = 300$

$y = \frac{100}{x^2}$

Surface area, A $= 3xx + 8xy +3xx + 8x(ky)$
$A = 6x^2 + (k+1) \frac{800}{x}$

$\frac{dA}{dx} = 12x - (k+1) \frac{800}{x^2} = 0$

$12x^3 = (k+1) 800$

$x^3 = \frac{200(k+1)}{3}$

$x = [\frac{200(k+1)}{3}]^{\frac{1}{3}}$

$\frac{d^2A}{dx^2} = 12 + (k+1) \frac{1600}{x^3}$

$= 12 + (k+1) \frac{1600}{\frac{200(k+1)}{3}}$

$= 36 > 0$

$\therefore, x = [\frac{200(k+1)}{3}]^{\frac{1}{3}}$ gives the minimum total surface area.

(ii)

$\frac{y}{x} = \frac{100}{x^2}$

$\frac{y}{x} = \frac{100}{\frac{200(k+1)}{3}} = \frac{3}{2(k+1)}$

(iii)
Since $0 k \le 1$

$1 < k+1 \le 2$

$\frac{1}{2} \le \frac{1}{k+1} < 1$

$\frac{3}{4} \le \frac{3}{2(k+1)} < \frac{3}{2}$

$\frac{3}{4} \le \frac{y}{x} < \frac{3}{2}$

(iv)
Since box has square ends, $x = y$

$1 = \frac{3}{2(k+1)}$

$k = \frac{1}{2}$

KS Comments:

As usual with all minimum or maximum problem, we need to show the second order derivative. (iii) takes a bit of skills but students who take the first step and try to write something at least will probably find the answer. (iv) requires students to interpret what it means to have square ends.

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KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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2010 A-level H2 Mathematics (9740) Paper 1 Question 9 Suggested Solutions

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