**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 $50,50$