If x y=4 and x

Question

If x y=4 and x<0 then maximum value of x+16 y is- (A) 8 (B) -8 (C) 16 (D) -16

Tardigrade Admin · Accepted Answer

x y=4, x<0 Let S=x+16 y S=x+64 / x <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/89cd60ff37ad03b34b3b30304154c2f3-.png" /> S=(x2+64/x) (d S/d x)=((x-8)(x+8)/x2) S is max at x=-8 ∴ S max =-16

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

8

$- 8$

16

$- 16$

$x y = 4, x < 0$
Let $S = x + 16 y$
$S = x + 64/ x$

$S = \frac{x ^{2} + 64}{x}$
$\frac{d S}{d x} = \frac{( x - 8 ) ( x + 8 )}{x ^{2}}$
$S$ is $max$ at $x = - 8$
$∴ S_{m a x} = - 16$

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

8

−8

16

−16

xy=4,x<0 Let S=x+16y S=x+64/x S=xx2+64​ dxdS​=x2(x−8)(x+8)​ S is max at x=−8 ∴Smax​=−16

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

$- 8$

$- 16$

$x y = 4, x < 0$
Let $S = x + 16 y$
$S = x + 64/ x$

$S = \frac{x ^{2} + 64}{x}$
$\frac{d S}{d x} = \frac{( x - 8 ) ( x + 8 )}{x ^{2}}$
$S$ is $max$ at $x = - 8$
$∴ S_{m a x} = - 16$