If the solve the problem

(b) 16

Given : $x+y=8$

$\Rightarrow y=8-x$           ....(1)

Let $f(x)$ be $x y$.

$\Rightarrow f(x)=x(8-x)$                       $[$ From eq. $(1)]$

$\Rightarrow f^{\prime}(x)=8-2 x$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 8-2 x=0$

$\Rightarrow 8=2 x$

$\Rightarrow x=4$

$\Rightarrow y=8-4=4$           $[$ From eq. $(1)]$

Now,

$f^{\prime \prime}(x)=-2$

$\Rightarrow f^{\prime \prime}(4)=-2<0$

So, $x=4$ is a local maxima.

Hence, the local maximum value is given by

$f(4)=4 \times 4=16$