Explanation

(d)
Given, $y = \frac{8 - x}{2}$
Let $p = xy = \frac{8x - x^{2}}{2}$
$\Rightarrow \ \ \ \ \ \ \ \frac{dp}{dx} = \frac{1}{2}(8 - 2x)$
For maxima or minima, put $\frac{dp}{dx} = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ 8 - 2x = 0$
$\Rightarrow \ \ \ \ \ \ x = 4$
Again, $\frac{d^{2}p}{dx^{2}} = - 1$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \left( \frac{d^{2}p}{dx^{2}} \right) = - 1 < 0$
Thus, function is maximum at $x = 4$ and $y = 2$
Therefore, maximum value of $p = 4 \times 2 = 8$