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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

$\begin{array}{1 1} 8\;and\;8 \\ 9 \;and\;7 \\ 10\;and\;6 \\ 11 \;and \;5 \end{array} $

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Toolbox:
  • $\large\frac{d}{dx}$$(x.y)=x.\large\frac{d}{dx}$$(y)+y.\large\frac{d}{dx}$$(x)$
Given
Sum of the No. = 16
Let sum of cubes be S
Let two numbers be $x$ and $16-x$
Sum of cubes $S=x^3+(16-x)^3$
Differentiating with respect to x we get
$\large\frac{dS}{dx}$$=3x^2+3(16-x)^2.(-1)$
$\quad\;\;=3x^2-3(16-x)^2.(-1)$
$\quad\;\;=3(32x-256)$
Step 2:
For minimum value
$\large\frac{dS}{dx}$$=0$
$3(32x-256)=0$
$x=\large\frac{256}{32}$$=8$
Also $\large\frac{d^2S}{dx^2}$$=96>0$
$\Rightarrow S$ is minimum at $x=8$
Hence the required numbers are $8$ & $(16-8)\Rightarrow$ 8 & 8
answered Aug 8, 2013 by sreemathi.v
edited Aug 20, 2013 by sharmaaparna1
 

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