$x_0 = \max(x_1, x_2, \ldots, x_{12})$
$x_0$ will be minimum if $(x_1, x_2, \ldots, x_{12})$ are close to each other.
$\frac{100}{12} = 8.33$
Therefore, $(\max(x_1, x_2, \ldots, x_{12}))$ will be minimum if $((x_1, x_2, \ldots, x_{12})$ = (9, 9, 9, 9, 8, 8, 8, 8, 8, 8, 8, 8)).
Hence, Option D is correct.
