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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find two numbers whose sum is 24 and whose product is as large as possible.

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Toolbox:
  • $\large\frac{d}{dx}\big(x^n)=nx^{n-1}$
  • For maximum value $\large\frac{dP}{dx}$$=0$
Step 1:
Given
Sum of two no$\rightarrow$ 24
Product $\rightarrow$ as large as possible
Let the required number be $x$ and $(24-x)$
Their product $P=x(24-x)$
$\qquad\qquad\quad\;\;\;=24x-x^2$
Step 2:
For maximum value
Now $\large\frac{dP}{dx}$$=0$
$P=24x-x^2$
Differentiating with respect to x we get
$\large\frac{dP}{dx}$$=24-2x$
$24-2x=0$
$-2x=-24$
$x=12$
$\large\frac{d^2P}{dx^2}$$=-2 < 0$
$\Rightarrow P$ is minimum at $x=12$
Hence the required numbers are 12 and $(24-12)\Rightarrow (i.e)12$
answered Aug 8, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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