The sum of two positive numbers is 16. What is the smallest value of the sum of their squares?

Solution:

Let x be 16 - x be the required numbers.

Let S be the sum of squares of these values,

S = x² + (16 - x)²

By solving the above equation using algebraic identity (a - b)², we get

S = x² + 256 - 32x + x²

= 2x² - 32x + 256

Differentiate w.r.t x

dS/dx = 4x - 32

Differentiate again w.r.t x

d²S/ dx² = 4

Condition for S to have minimum value is dS/dx = 0 and d²S/dx² > 0

Consider dS/dx = 0

4x - 32 = 0

x = 8

When x = 8, d²S/dx² = 4 > 0

S is minimum when x = 8

S = Value of S at x =8

= 8² + (16 - 8)²

= 8²+ 8²

= 128

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The sum of two positive numbers is 16. What is the smallest value of the sum of their squares?

Summary:

The sum of two positive numbers is 16, the smallest value of the sum of their squares is 128.