Explanation

(d)
Let $A = x^{2}y = x^{2}(8 - x)$ [given, $x + y = 8$]
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ A = 8x^{2} - x^{3}$
$\Rightarrow \ \ \ \ \frac{dA}{dx} = 16x - 3x^{2}$
For maxima or minima, put $\frac{dA}{dx} = 0$
$\Rightarrow \ \ \ \ \ 16x - 3x^{2} = 0\ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ x = 0,\frac{16}{3}$
Now, $\frac{d^{2}A}{dx^{2}} = 16 - 6x$
$\left( \frac{d^{2}A}{dx^{2}} \right)_{x = \frac{16}{3}} = 16 - 32 < 0$, maxima
$\therefore$ Maximum value$= 8\left( \frac{16}{3} \right)^{2} - \left( \frac{16}{3} \right)^{3} = \frac{2048}{27}$