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!