Show that the surface area of a closed cuboid with square base an… - Vidyadip

Question

Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.

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Answer

Let the sides of cuboid be x, x, y
$\Rightarrow \text{x}^{2} \text{y} = \text{k and S = 2} \text{(x}^{2} + \text{xy + xy)} = 2 (\text{x}^{2} + \text{2xy})$
$\therefore \text{S} = 2 \bigg[\text{x}^{2} + \text{2x} \frac{\text{k}}{\text{x}^{2}}\bigg] = 2\bigg[\text{x}^{2} + \frac{\text{2k}}{\text{x}}\bigg]$
$\frac{\text{ds}}{\text{dx}} = 2 \bigg[\text{2x} - \frac{\text{2k}}{\text{x}^{2}}\bigg]$
$\therefore\frac{\text{ds}}{\text{dx}} = 0 \Rightarrow \text{x}^{3} = \text{k} = \text{x}^{2}\text{y} \Rightarrow \text{x = y}$
$\frac{\text{d}^{2}\text{s}}{\text{dx}^{2}} = 2 \bigg[2 + \frac{\text{4k}}{\text{x}^{3}}\bigg]> \text{ }\therefore \text{ }\text{x = y}$will given minimum surface area.
and x = y, means sides are equal.
$\therefore$ Cube will have minimum surface area.

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