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

(i)

Volume, V

Surface area, A

gives the minimum total surface area.

(ii)

(iii)

Since

$latex 1 < k+1 \le 2 $ $latex \frac{1}{2} \le \frac{1}{k+1} < 1$ $latex \frac{3}{4} \le \frac{3}{2(k+1)} < \frac{3}{2}$ $latex \frac{3}{4} \le \frac{y}{x} < \frac{3}{2}$ (iv) Since box has square ends, $latex x = y$ $latex 1 = \frac{3}{2(k+1)}$ $latex k = \frac{1}{2}$

### 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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