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(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 $0 k \le 1$

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