CUET Preparation Today
Two positive numbers x and y such that $x+y = 60$ and $xy^3$ is maximum are : |
$x=30, y= 30 $ $x=20, y= 40 $ $x=15, y= 45 $ $x=55, y= 5 $ |
$x=15, y= 45 $ |
Given $x+y=60$ Let $f(x)=xy^{3}$ $y=60-x$ $f(x)=x(60-x)^{3}$ $f'(x)=(60-x)^{3}+x\cdot3(60-x)^{2}(-1)$ $=(60-x)^{2}\left[(60-x)-3x\right]$ $=(60-x)^{2}(60-4x)$ $f'(x)=0$ $60-4x=0$ $x=15$ $y=60-15=45$ The two numbers are $x=15$ and $y=45$. |
