The sum of two positive numbers is 12 and the product of one and the square of the other is maximum. Find the numbers.

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 one and the square of
the other is a maximum.

We'll write the product of integers
as:

P = x*y^2

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

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

We'll expand the
square:

p(x) = x(144 - 24x +
x^2)

We'll remove the
brackets:

p(x) = 144x - 24x^2 +
x^3

 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) = (144x - 24x^2 +
x^3)'

p'(x) = 144 - 48x +
3x^2

p'(x) = 0

144 - 48x +
3x^2 = 0

We'll divide by 3 and we'll apply symmetric
property:

x^2 - 16x + 48 =
0

We'll apply the quadratic
formula:

x1 =
[16+sqrt(256-192)]/2

x1 =
(16+8)/2

x1 = 12

x2 =
(16-8)/2

x2 =
4

The positive numbers are: x = 4 and y = 12
- 4 = 8.

Note:
We cannot choose x = 12 because x+y = 12, so if x = 12, y =
0.

If y = 0, the condition x*y^2 is a maximum is impossible
because x*0 = 0 and 0 is not the root for p'(x) = 0.