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Q. If $x y=4$ and $x<0$ then maximum value of $x+16 y$ is-

Application of Derivatives

Solution:

$x y=4, x<0$
Let $ S=x+16 y$
$S=x+64 / x$
image
$ S=\frac{x^2+64}{x} $
$ \frac{d S}{d x}=\frac{(x-8)(x+8)}{x^2}$
$S$ is $\max$ at $x=-8$
$\therefore S_{\max }=-16$