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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.
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