Two positive numbers x and y whose sum is 16 and xy^3 is maximum are :

The correct answer is Option (3) → $x=4, y=12$

$x+y=16$

$⇒x=16-y$   ...(1)

$P=x.y^3$

$=(16-y)y^3$

$=16y^3-y^4$

$\frac{dP}{dy}=48y^2-4y^3$

$⇒48y^2-4y^3=0$

$⇒4y^2(12-y)=0$

$y=0$ or $y=12$

Substituting $y=12$ in equation (1),

$x=4$