Find three positive numbers whose sum is 12 and the sum of whose squares is as small as possible.

Solution: 

Let the positive numbers be x, y and z.

Given that the sum of numbers is 12.

⇒ x + y + z = 12 --- (1)

⇒ z = 12 - x - y

Let the sum of the squares of the number be S.

∴ S = x² + y² + z²

⇒ S = x² + y² + (12 - x - y)² --- (2)

To find the minimum value of S, we will optimize the function and differentiate it partially with respect to x and y and make dS/dx = 0

Partially differentiating (2) w.r.t x

2x + 2(12 - x - y)(-1) = 0

2x - 24 + 2x + 2y = 0

4x + 2y - 24 = 0

y = 12 - 2x --- (3)

Partially differentiating equation (2) w.r.t y,

2y + 2(12 - x - y)(-1) = 0

2y - 24 + 2x + 2y = 0

4y + 2x - 24 = 0

x = 12 - 2y --- (4)

By substituting the value of y = 12 - 2x in equation (4),

x = 12 - 2(12 - 2x)

x = 12 - 24 + 4x

4x - x = 24 - 12

3x = 12

⇒ x = 4

By substituting x = 4 in equation (3)

y = 12 - 2(4)

y = 12 - 8

⇒ y = 4

By substituting the values of x = 4 and y = 4 in equation (1), we get,

4 + 4 + z = 12

z = 12 - 8 = 4

From S = x² + y² + z², we get

S = 4² + 4² + 4²

S = 48

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Find three positive numbers whose sum is 12 and the sum of whose squares is as small as possible.

Summary:

The three positive numbers whose sum is 12 are (x, y, x) = (4, 4, 4) and the sum of their squares is 48.