CUET Preparation Today
Two positive numbers x and y whose sum is 16 and $xy^3$ is maximum are : |
$x=5, y=11$ $x=7, y=9$ $x=4, y=12$ $x=2, y=10$ |
$x=4, y=12$ |
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), $x=4$ |
