Find two numbers whose sum is 9 if the product of one by the square of the other is a maximum

x+y = 9

the product z is
z = xy^2 = x(9-x)^2 = 81x - 18x^2 + x^3
dz/dx = 81 - 36x + 3x^2 = 3(x^2-12x+27)
dz/dx=0 when x = 3 or 9
Pick x=3, so y=6

To find two numbers whose sum is 9 and maximize the product of one number multiplied by the square of the other, we can use calculus.

Let's assume the two numbers are x and 9 - x.

We need to maximize the function f(x) = x(9 - x)^2.

To find the maximum of this function, we can take the derivative:

f'(x) = (9 - x)^2 - 2x(9 - x)

To find critical points, set f'(x) = 0:

(9 - x)^2 - 2x(9 - x) = 0

Expanding and simplifying:

x^2 - 9x + 18 = 0

Factoring the equation:

(x - 6)(x - 3) = 0

So, x = 6 or x = 3.

To determine which x value gives us the maximum, we can evaluate the second derivative f''(x):

f''(x) = 2 - 4(9 - x)

Let's plug in x = 6:

f''(6) = 2 - 4(9 - 6) = 2 - 12 = -10

Since f''(6) is negative, it means x = 6 gives us a maximum.

Therefore, the two numbers are 6 and 9 - 6 = 3.

Hence, the two numbers whose sum is 9 and maximize the product of one number multiplied by the square of the other are 6 and 3.

To find two numbers whose sum is 9, let's assume the two numbers are x and 9 - x.

Given that the product of one number by the square of the other is a maximum, we can express this as a function:

f(x) = x(9 - x)^2

To find the maximum value of this function, we can differentiate it with respect to x and set the derivative equal to zero:

f'(x) = 0

Let's calculate the derivative of f(x):

f'(x) = 0 - (9 - x)^2 + x * 2 * (9 - x) * (-1)

Simplifying further:

f'(x) = - (9 - x)^2 - 2x(9 - x)

Now, set the derivative equal to zero and solve for x:

0 = - (9 - x)^2 - 2x(9 - x)

Expanding and simplifying:

0 = - (81 - 18x + x^2) - (18x - 2x^2)

0 = - 81 + 18x - x^2 - 18x + 2x^2

0 = x^2 - 9

Solving for x using the quadratic formula:

x^2 - 9 = 0

(x - 3)(x + 3) = 0

So, x can be either 3 or -3.

Therefore, the two numbers whose sum is 9 are (3, 6) or (-3, 12).

Well, these numbers must really enjoy hanging out together because their sum is 9! Let's call the numbers x and y. So we have the equation:

x + y = 9

Now, the goal is to find the maximum value of the product of one number with the square of the other. Let's call this product P. So we have:

P = x * y^2

To find the maximum value, we can employ a bit of mathematical circus magic. This is where the fun part comes in! Let's solve the first equation for x:

x = 9 - y

Now we substitute this expression into the equation for P:

P = (9 - y) * y^2

To find the maximum, let's bring in some acrobatics! We need to find the critical points of P, which means taking the derivative of P with respect to y and setting it equal to zero. Let the tricks begin!

dP/dy = -y^2 + 18y - 9

Setting this equal to zero:

-y^2 + 18y - 9 = 0

Now, let's use quadratic formula and hope the answer doesn't turn into a clownfish!

y = (-b ± √(b^2 - 4ac)) / (2a)

With a = -1, b = 18, and c = -9, let's do some juggling!

y = (-18 ± √(18^2 - 4(-1)(-9))) / (2(-1))

After a whirlwind of calculations...

y = 4.5 or 1.5

Now, let's substitute these values back into the equation for x:

For y = 4.5, x = 9 - 4.5 = 4.5

For y = 1.5, x = 9 - 1.5 = 7.5

Voila! We have two pairs of numbers whose sum is 9 and have the maximum product of one number multiplied by the square of the other: (4.5, 4.5) and (7.5, 1.5).

I hope you enjoyed this mathematical circus act!