$\begin{array}{1 1} 35,25 \\ 30 ,30 \\ 15,45 \\ 20,40 \end{array} $

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- $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$

Step 1:

Given

We have $x+y=60$

$y=60-x$

Let $P=xy^3$

$\qquad=x(60-x)^3$

On differentiating with respect to x we get

$\large\frac{dP}{dx}$$=x.3(10-x)^2(-1)+(60-x)^3.1$

$\quad\;\;=-3x(60-x)^2+(60-x)^3$

$\quad\;\;=(60-x)^2[-3x+60-x]$

$\quad\;\;=(60-x)^2[60-4x]$

Step 2:

Now $\large\frac{dP}{dx}$$=0$

$\Rightarrow (60-x)^2(60-4x)=0$

$\therefore$Either $(60-x)^2=0$

(i.e)$x=60$ or

$60-4x=0$

$-4x=-60$

$x=15$

The value $x=60$ is rejected as it makes $y=0$ .

Step 3:

At $x=15$

When $x$ is slightly $<15\large\frac{dP}{dx}$$=(+)(+)=+ve$

When $x$ is slightly $>15\large\frac{dP}{dx}$$=(+)(-)=-ve$

$\Rightarrow \large\frac{dP}{dx}$ changes sign from +ve to -ve as $x$ increases through 15.

$\therefore P$ is maximum at $x=15$

Hence the required numbers are 15 and $(60-15)\Rightarrow 45$

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