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# Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.

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

Toolbox:
• $\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}$$=0 \Rightarrow (60-x)^2(60-4x)=0 \thereforeEither (60-x)^2=0 (i.e)x=60 or 60-4x=0 -4x=-60 x=15 The value x=60 is rejected as it makes y=0 . Step 3: At x=15 When x is slightly <15\large\frac{dP}{dx}$$=(+)(+)=+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 $(60-15)\Rightarrow 45$
edited Aug 20, 2013