The given condition is:
$x+y=8 \Rightarrow y=8-x$
and let, the product of $\mathrm{x}$ and $\mathrm{y}$ be, $\mathrm{P}=\mathrm{xy}$
$\Rightarrow P=x(8-x)=8 x-x^{2}$
Differentiating w.r.t. $\mathrm{x}$
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dx}}=8-2 \mathrm{x}$
and $\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}=-2 < 0$
Put $\frac{\mathrm{dP}}{\mathrm{dx}}=0$, for maxima or minima
$8-2 x=0$
$\Rightarrow x=\frac{8}{2}=4$ and $y=4$
and $\left(\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}\right) < 0$ when $\mathrm{x}=\mathrm{y}$
$\therefore$ P is maximumat $\mathrm{x}=4$ Maximum value of $\mathrm{P}=4 \cdot(8-4)=4.4=16$