Question

Find two positive numbers $x \space and\space y$ such that $x+y=60$ and $xy^{3}$ is maximum.

Answer 1

The two numbers are $x \space and\space y$ such that$x+y=60.$

$∴ y= 60-x$

$Let f(x) = xy^{3}.$

$f(x)= x(60-x)^{3}$

$f'(x)=(60-x)^{3}$

$f'(60-x)^{2}[60-x-3x]$

$=(60-x)^{2}(60-4x)$

And,$f''(x)=-2(60-x)-4(60-x)^{2}$

$=-2(60-x)[60-4x+2(60-x)]$

$=-2(60-x)(180-6x)$

$=-12(60-x)(30-x)$

Now$,f'(x)=0=x=60 \space or\space x=15$

$When \space x=60,f''(x)=0.$

$When\space x=15,f''(x)=-12(60-15)(30-15)=-12\times45\times15<0.$

∴By second derivative test,$x = 15$ is a point of local maxima of$f.$

Thus, function $xy^{3}$ is maximum when $x = 15$ and $y = 60−15=45$. Hence, the required numbers are $15 \space and \space45.$