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Let the number be x and y;
Given that the sum is 36
==> x + y= 36
We will write as function of y:
==> y= 36-x .............(1)
Now we need to find the numbers such that their product is a maximum.
Let P be the product:
==> P = x*y
But y= (36-x)
==> P = x*(36-x)
==> P = 36x - x^2
Now we need to find the maximum point of P
Since the sign of x^2 is negative, then the function has a maximum.
==> P' = 36 - 2x
==> 36- 2x = 0
==> 2x= 36
==> x= 18
==> y= 36-18 =18
Then the numbers are 18 and 18 and the maximum product is:
\p = 18*18 = 324
Since the sum of the two numbers is 36, we cal assume the numbers to be x and 36-x.
Let the product p(x) , of these two numbers x and 36-x.
Then p(x) = x(36-x) = 36x-x^2.
P(x) = 36x=x^2 is maximum , for x = c, for which p'(c) = 0 and P"(c) > 0.
So we differentiate P(x) and set P'(x) = 0 and find the the solution of P'(x) = 0. And examine if P"(c) is < 0.
P'(x) = (36x-x^2)' = 36 -2x
P'(x) = 0 gives : 36 = 2x. Or x = c = 36/2 = 12.
P"(x) = (-2x) = -2 < 0 for all x.
Therefore P(x) = 36x-x^2 when x= 18.
P(x) = 18(36-18) = 18^2 = 324 is the maxumum product.
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