# 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 = x^{2} + (16 - x)^{2}

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

S = x^{2} + 256 - 32x + x^{2}

= 2x^{2} - 32x + 256

Differentiate w.r.t x

dS/dx = 4x - 32

Differentiate again w.r.t x

d^{2}S/ dx^{2} = 4

Condition for S to have minimum value is dS/dx = 0 and d^{2}S/dx^{2} > 0

Consider dS/dx = 0

4x - 32 = 0

x = 8

When x = 8, d^{2}S/dx^{2} = 4 > 0

S is minimum when x = 8

S = Value of S at x =8

= 8^{2} + (16 - 8)^{2}

= 8^{2}+ 8^{2}

= 128

## 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.

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