Find two positive numbers

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

$\Rightarrow y=60-x$

Let $f(x)=x y^{3}$

$\begin{aligned} \Rightarrow f(x) &=x(60-x)^{3} \\ \therefore f^{\prime}(x) &=(60-x)^{3}-3 x(60-x)^{2} \\ &=(60-x)^{2}[60-x-3 x] \\ &=(60-x)^{2}(60-4 x) \end{aligned}$

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

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

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

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

Now, $f^{\prime}(x)=0 \Rightarrow x=60$ or $x=15$\

When $x=60, f^{\prime \prime}(x)=0$

When $x=15, f^{\prime \prime}(x)=-12(60-15)(30-15)=-12 \times 45 \times 15<0$.

$\therefore$ By second derivative test, $x=15$ is a point of local maxima of $f$. Thus, function $x y^{3}$ is maximum when $x=15$ and $y=60-15=45$.

Hence, the required numbers are 15 and 45.