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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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} $

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1 Answer

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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}$$=0$
$\Rightarrow (60-x)^2(60-4x)=0$
$\therefore$Either $(60-x)^2=0$
(i.e)$x=60$ or
$60-4x=0$
$-4x=-60$
$x=15$
The value $x=60$ is rejected as it makes $y=0$ .
Step 3:
At $x=15$
When $x$ is slightly $<15\large\frac{dP}{dx}$$=(+)(+)=+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 $(60-15)\Rightarrow 45$
answered Aug 8, 2013 by sreemathi.v
edited Aug 20, 2013 by sharmaaparna1
 

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