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

*print*Print*list*Cite

### 2 Answers

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.

Let the two numbers whose sum is 12 be x and 12-x.

Then the product of one (x) and the square of the other (12-x)^2 is P(x) x (12-x)^2.

Now P(x) is maximum, when P'(c) = 0 for which P"(c) < 0.

So we find P'(x) and equate to zero and solve for x = c. Then see whether P"(c) is less than zero.

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

P'(x) = {x(144-24x+x^2}'

P'(x) = {144x -24x^2+x^3}'

P'(x) = 144-24*2x +3x^2.

Equate t0 zero:

144-48x +3x^2 = 0. Divide by 3.

48 -16x+x^2 = 0

(x-12)(x-4) = 0.

x-12 = 0 or x = 4.

P"(x) = {144-48x+3x^2}'

P"(x) = -48+6x .

P(4) = -48+6*4 = -24 < 0.

Therefore P(4) = 4(12-4)^2 = 256 is the maximum