Find two positive numbers whose sum is 50 such that the sum of their squares is minimum?

let one number be x

then the other is 50-x

let the sum of their squares be S
S = x^2 + (50-x)^2
= 2x^2 - 100x + 2500
dS/dx = 4x - 100
= 0 for a min of S
4x-100 = 0
x = 25

The numbers are 25 and 25

Well, if we're looking for two positive numbers whose sum is 50, and we want the sum of their squares to be as small as possible, let's call the two numbers x and y.

To minimize the sum of their squares, we need to make sure that x and y are as close as possible to each other. So, let's try splitting 50 in half and choose x = 25 and y = 25.

Now, if we calculate the sum of their squares, we get 25^2 + 25^2 = 625 + 625 = 1250.

So, there you have it! The two positive numbers whose sum is 50, with the smallest sum of their squares possible, are 25 and 25.

To find the two positive numbers whose sum is 50 such that the sum of their squares is minimum, we can use calculus. Let's call the two numbers x and y.

1. Write the equation: x + y = 50.
2. We need to find the minimum value of the sum of their squares, which means we can minimize the function f(x) = x^2 + y^2.
3. Solve the equation for y: y = 50 - x.
4. Substitute y in the function: f(x) = x^2 + (50 - x)^2.
5. Expand the equation: f(x) = x^2 + (2500 - 100x + x^2).
6. Simplify: f(x) = 2x^2 - 100x + 2500.
7. To find the minimum value, take the derivative of f(x) and set it to 0: f'(x) = 4x - 100 = 0.
8. Solve for x: 4x = 100, x = 25.
9. From the initial equation, find y: y = 50 - x = 50 - 25 = 25.

Therefore, the two positive numbers whose sum is 50 and the sum of their squares is minimum are 25 and 25.

To find two positive numbers whose sum is 50 such that the sum of their squares is minimum, we can use optimization techniques involving calculus and equations.

Let's assume that the two numbers are x and y, with x and y both being positive. We can then set up the following equations:

1. The sum of the two numbers: x + y = 50 (as given)
2. The sum of their squares: f(x, y) = x^2 + y^2

To find the minimum value of f(x, y), we need to find the critical points. We can use the method of Lagrange multipliers to solve this optimization problem.

First, we rewrite equation 1: y = 50 - x.

Substituting this into equation 2, we get f(x) = x^2 + (50 - x)^2.

Now, let's differentiate f(x) with respect to x and set it equal to zero to find the critical point(s):

f'(x) = 2x - 2(50 - x) = 0

Simplifying, we have 2x - 100 + 2x = 0.

Combining like terms, we get 4x - 100 = 0.

Solving for x, we have x = 25.

Substituting this value of x back into equation 1, we find y = 50 - x = 50 - 25 = 25.

Therefore, the two positive numbers whose sum is 50 and the sum of their squares is minimum are x = 25 and y = 25.