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Find two numbers whose sum is $100 $ and whose product is a maximum.

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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 numbers be x and 100 - x (since their sum is 100)
Step 2:
Their product is $p= x(100-x)$
$\qquad=100x -x^2$
Step 3:
$\large\frac{dp}{dx}$$=100-2x$
At extreme values of P, $\large\frac{dp}{dx}$$=0 =>100-2x=0$
$\qquad=>x=50$
$\large\frac{d^2p}{dx^2}$$=-2 < 0=>x=50 $ corresponds to a local maximum.
Step 4:
$\therefore p$ is maximum when $x=50$
The two numbers are $50,50$

 

answered Aug 1, 2013 by meena.p
 

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