Given that, $ y=\frac{8-x}{2} $
Let $ p=xy=x\left( \frac{8-x}{2} \right) $
$ =\frac{8x-{{x}^{2}}}{2} $
On differentiating w.r.t. $ x, $ we get
$ \frac{dp}{dx}=\frac{1}{2}(8-2x) $
Put $ \frac{dp}{dx}=0 $ for maxima or minima
$ \therefore $ $ 8-2x=0 $
$ \Rightarrow $ $ x=4 $
Again $ \frac{{{d}^{2}}p}{d{{x}^{2}}}=\frac{1}{2}(-2)=-1 $
$ {{\left( \frac{{{d}^{2}}p}{d{{x}^{2}}} \right)}_{x=4}}=-1<0 $
Thus, function is maximum at
$ x=4 $ and $ y=2 $
Therefore, maximum value of $ p=4\times 2=8 $ .