Let us assume one of the numbers is x. As the sum of the two numbers is 20, the other number is 20-x

The product of the two numbers is a function f(x) = x*(20-x) = 20x- x^2.

Now we need to find the maximum value of f (x). For this we need the derivative of f(x) and have to equate it to 0.

f’(x) = 20 – 2x = 0

=> 10 – x =0

=> x =10.

Now f’’(x) = -2 which is negative at x=10. So f (10) is truly the maximum value.

If x=10, the first number is 10 and the second number is 20-10 = 10.

**So the two required numbers are 10 and 10.**

Let the numbers whose sum is 20 be x and 20-x.

Then the product of the nimbers x and 20-x is P(x) = x(20-x).

P(x) = 20x-x^2

For the maximum , P'(c) = 0 and p"(c) <0.

P'(x) = ((20x-x^2)' = 20-2x

P'(c) = 0 for 20-2x = 0. Or for x =10.

P"(x) = (20-2x)' = -2. So P'(10 ) -2 which is less than zero.

Therefore for x = 10, the product P(x) = 10(20-10) = 100 is the maximum.