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Find two positive numbers whose product is $100$ and whose sum is minimum.

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Toolbox:
  • Second derivative test: Suppose $f$ is continuous on an open interval that contains
  • (i)If $f'(c)=0\; and\; f''(c) >0$ then $f$ has a local minimum at $c$.
  • (ii)If $f'(c)=0\;and \; f''(c)<0$ then $f$ has a local maximum at $c$
Step 1:
Let the two positive numbers be x and $\large\frac{100}{ x}$ (since their product is 100)
Step 2:
The sum of two numbers $s= x +\large\frac{100}{x}$
$\large\frac{ds}{dx}=1-\large\frac{100}{x^2}$
At extreme values of $s, \large\frac{ds}{dx}$$=0=>1-\large\frac{100}{x^2}$$=0$
$=>x^2=100\;or\; x=10$
Now $\large\frac{d^2s}{dx^2}=\frac{200}{x^3}$
$\large\frac{d^2s}{dx^2(x=10)}=\frac{200}{1000}=\frac{1}{5} > 0$
$\therefore x=10$ corresponds to a minimum value of S.
The two numbers are 10 and 10
answered Aug 1, 2013 by meena.p
 

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