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Total Volume $= 3x(x)y = 300$

$x^2y = 100$

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

Total Surface Area, $A = 3x(x) + 8x(y) + 3x(x) + 8x(ky)$

$= 6x^2 + (k+1) \frac{800}{x}$

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

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

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

$\frac{d^2A}{dx^2} = 12 + (k+1) \frac{1600}{x^3} = 12 + (k+1)\frac{1600}{\frac{200}{3}(k+1)} = 36 > 0$

Thus, $x=^3 \sqrt{\frac{200}{3}(k+1)}$ gives a minimum total surface area.

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

(iii)
$0 \textless k \le 1$

$1 \textless k+1 \le 2$

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

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

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

(iv)
If the box has square ends, then $y=x$

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

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