Find two positive numbers whose sum is 8 such hat when the cube of th first number is multiplied by the second number, the result is maximum.
the numbers are x and 8-x, so
f(x) = x^3 (8-x)
f' = 4x^2 (6-x)
f'=0 at x=0,6
and has a maximum at x=6
To find two positive numbers whose sum is 8 such that the product of the cube of the first number and the second number is maximum, we can use the concept of calculus and optimization.
Let's call the two positive numbers x and 8 - x (where x is the first number).
The product we want to maximize is P = x^3 * (8 - x).
To find the maximum value of P, we need to identify the critical points of the function P(x), which occur when the derivative of P with respect to x is equal to 0.
1. Calculate the derivative of P(x) with respect to x:
P'(x) = 3x^2 * (8 - x) - x^3
2. Set the derivative equal to 0 and solve for x:
3x^2 * (8 - x) - x^3 = 0
3. Simplify the equation:
24x - 3x^2 - x^3 = 0
4. Rearrange the equation:
x^3 - 3x^2 + 24x = 0
5. Factor out x:
x(x^2 - 3x + 24) = 0
6. Solve for x using either factoring or the quadratic formula:
x = 0 (which is not positive, so it can be discarded as a solution)
x^2 - 3x + 24 = 0
Using the quadratic formula, we get:
x = (3 ± √(3^2 - 4*24)) / 2
x ≈ 2.2643 (approximately)
Since x must be a positive number, the second number (8-x) ≈ 5.7357 (approximately).
So, the two positive numbers whose sum is 8 and maximize the product of the cube of the first number and the second number are approximately 2.2643 and 5.7357.