the sum of tow postive real number is 20
the minum possible values or the sum of the cube is
the answer is 2000 explain why
Let's denote the two positive real numbers as x and y.
According to the given information, x + y = 20.
To find the minimum possible value for the sum of the cubes of x and y, we can use the AM-GM inequality, which states that the arithmetic mean is always greater than or equal to the geometric mean.
In this case, the arithmetic mean of x and y is (x + y)/2, which is equal to 10.
The geometric mean of x and y is √(xy).
Now, using the AM-GM inequality, we have:
(√(xy))^3 ≤ ((x + y)/2)^3
Cubing both sides:
xy(x + y) ≤ (x + y)^3/8
Since x + y = 20, we have:
xy(20) ≤ 20^3/8
20xy ≤ 8000/8
20xy ≤ 1000
Dividing both sides by 20:
xy ≤ 50
We want to find the minimum possible value for the sum of the cubes, which is x^3 + y^3. Using the identity x^3 + y^3 = (x + y)(x^2 - xy + y^2), we can rewrite it as:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
Since x + y = 20, we have:
x^3 + y^3 = 20(x^2 - xy + y^2)
We want to minimize this expression, so we need to minimize the value of (x^2 - xy + y^2).
Using the result we found earlier, xy ≤ 50, we can substitute it into the expression:
x^3 + y^3 = 20(x^2 - xy + y^2) ≥ 20(x^2 - 50 + y^2)
Now, we need to minimize (x^2 - 50 + y^2). Since both x and y are positive real numbers, their squares will be positive as well. Therefore, the minimum possible value of (x^2 - 50 + y^2) occurs when both x and y are as small as possible.
If we let both x and y be equal to √50, we have:
(x^2 - 50 + y^2) = (√50^2 - 50 + √50^2) = (50 - 50 + 50) = 50
Therefore, the minimum possible value of (x^2 - 50 + y^2) is 50.
Substituting this back into the expression for x^3 + y^3:
x^3 + y^3 = 20(x^2 - xy + y^2) ≥ 20(50) = 1000
Therefore, the minimum possible value for the sum of the cubes of x and y is 1000, not 2000 as stated in the question.
To find the minimum possible value of the sum of the cubes of two positive real numbers, let's assume the two numbers are x and y.
Given that the sum of the two numbers is 20, we have the equation:
x + y = 20
Now, we want to find the minimum possible value of the sum of their cubes, which is x^3 + y^3.
To find the minimum value, we can use the AM-GM inequality, which states that the arithmetic mean of a set of positive numbers is always greater than or equal to the geometric mean of the same set of numbers.
In this case, the AM-GM inequality can be applied to the numbers x^3 and y^3:
(x^3 + y^3) / 2 ≥ [(x^3)*(y^3)]^(1/2)
Since we want to find the minimum value, we can replace the greater than or equal to sign with an equal sign:
(x^3 + y^3) / 2 = [(x^3)*(y^3)]^(1/2)
Simplifying this equation, we get:
(x^3 + y^3) = 2 * [(x^3)*(y^3)]^(1/2)
Now, let's solve for x and y using the equation x + y = 20 and substitute in the original equation to find the minimum value:
x = 20 - y
(x^3 + y^3) = 2 * [(x^3)*(y^3)]^(1/2)
[(20 - y)^3 + y^3] = 2 * [(20 - y)^3 * y^3]^(1/2)
Expanding the equation, we get:
(8000 - 120y + 6y^2 - y^3 + y^3) = 2 * [(20 - y)^3 * y^3]^(1/2)
8000 - 120y + 6y^2 = 2 * [(20 - y)^3 * y^3]^(1/2)
To find the minimum value, we differentiate the equation with respect to y and equate it to zero:
d/dy [8000 - 120y + 6y^2] = 0
-120 + 12y = 0
12y = 120
y = 10
Substituting this value of y back into the equation x + y = 20, we get:
x + 10 = 20
x = 10
Therefore, the minimum value of the sum of the cubes of two positive real numbers is x^3 + y^3 = 10^3 + 10^3 = 2000.
Hence, the answer is 2000.