We'll note the integers as x and y.

The sum of the integers is 12.

x + y = 12

y = 12 - x

We also know that the product of integers is a maximum.

We'll write the product of integers as:

P = x*y

We'll substitute y by (12-x) and we'll create the function p(x):

p(x) = x*(12-x)

We'll remove the brackets and we'll get:

p(x) = 12x - x^2

The function p(x) is a maximum when x is critical, that means that p'(x) = 0

We'll calculate the first derivative for p(x):

p'(x) = (12x - x^2)'

p'(x) = 12 - 2x

p'(x) = 0

12 - 2x = 0

We'll divide by 2:

6 - x = 0

We'll subtract 6 both sides:

-x = -6

We'll divide by -1:

x = 6

**So, x is the critical value and the integers are x = 6 and y = 6.**

The 2 integers whose sum = 12 and the product is maximum.

Let x and 12-x be the 2 intefers.

Then their product P(x) = x(12-x)

For the maximum of P(x) at c , P'(c) = 0 and P"(c) < 0 is the condition by calculus.

P'(x) = {x(12-x)}

P'(x) = {12x-x^2}'

p'(x) = 12-2x. Equating to zero, we get 12-2x = 0, for x = 12/2 = 6.

P"(x) = (12-2x)' = -2 which is negative.

Therefore P(6) = 6*(12-6) = 36 is the maximum.