Minimizing Sum: Find two positive numbers whose sum is 100 and the sum of whose squares is a minimum.
let one number be x
let the other be 100-x
let the sum of their squares be S
S = x^2 + (100-x)^2
find dS/dx
set that equal to zero and solve for x
To find two positive numbers whose sum is 100 and the sum of whose squares is a minimum, we can use calculus techniques.
Let's give these two positive numbers names - let's call them x and 100-x. Therefore, x is the first number, and 100-x is the second number.
Now we need to find the sum of their squares, which we'll call S. We can express S as a function of x:
S(x) = x^2 + (100-x)^2
To find the minimum value of S(x), we can take the derivative of S(x) with respect to x and set it equal to zero:
dS/dx = 2x - 2(100-x)
Setting this equal to zero:
2x - 2(100-x) = 0
Simplifying:
2x - 200 + 2x = 0
Combining like terms:
4x - 200 = 0
Rearranging:
4x = 200
x = 50
So, x = 50 is the value that minimizes the sum of squares.
Substituting x = 50 back into the expression for S(x):
S(50) = 50^2 + (100-50)^2
Simplifying:
S(50) = 2500 + 2500
S(50) = 5000
Therefore, the minimum sum of squares is 5000, and the two positive numbers that add up to 100 with this minimum sum are 50 and 50.
To solve this problem, we can use a well-known mathematical concept called the AM-GM inequality. According to the AM-GM inequality, the arithmetic mean of a set of positive numbers is always greater than or equal to their geometric mean, with equality only when all the numbers are equal.
In this case, we are trying to minimize the sum of squares of two positive numbers, x and y, with the condition that their sum is 100.
Step 1: Set up the equations:
Let x and y be the two positive numbers.
We have the following two equations:
x + y = 100 (Equation 1)
x² + y² = minimum (Equation 2)
Step 2: Apply the AM-GM inequality:
According to the AM-GM inequality, we have:
(x² + y²)/2 ≥ √(x²y²) (Equation 3)
Step 3: Simplify Equation 3:
The left side of Equation 3 is the average of x² and y².
The right side of Equation 3 is the geometric mean of x² and y².
Step 4: Solve Equation 3:
Simplifying Equation 3, we get:
(x² + y²)/2 ≥ √(x²y²)
(x² + y²)/2 ≥ xy (Equation 4)
Step 5: Substitute Equation 1 into Equation 4:
Substituting Equation 1 into Equation 4, we get:
(100² - 2xy)/2 ≥ xy
(100² - 2xy) ≥ 2xy
100² ≥ 4xy
2500 ≥ xy (Equation 5)
Step 6: Find the minimum value of xy:
From Equation 5, we know that xy is less than or equal to 2500.
Step 7: Find the minimum values of x and y:
To find the minimum values of x and y, we can use the equality case of the AM-GM inequality. In this case, the numbers x and y will be equal to each other.
So, x = y = √(xy)
Since xy is less than or equal to 2500, the maximum possible value for x and y is when xy is equal to 2500.
Hence, x = y = √2500 = 50.
Step 8: Verify the solution:
To verify the solution, check if the sum of x and y is 100:
x + y = 50 + 50 = 100. This condition is satisfied.
Also, check if the sum of squares is minimum:
x² + y² = 50² + 50² = 2500 + 2500 = 5000. This is the minimum value.
Therefore, the two positive numbers that satisfy the given conditions and minimize the sum of squares are 50 and 50.