Answer:
The smallest possible value of the sum of their squares is 128.
Step-by-step explanation:
Let the first number = x
let the the second number = y
x + y = 16
y = 16 -x
minimum value, f = x² + y²
f(x) = x² + (16-x)²
f(x) = x² + 256 - 32x + x²
f(x) = 2x² - 32x + 256
f(x)' = 4x - 32
4x - 32 = 0
4x = 32
x = 32 / 4
x = 8
The absolute minimum value of x = 8
y = 16 - x
y = 8
The smallest possible value of the sum of their squares is given by x² + y²;
= (8)² + (8)²
= 128
Therefore, the smallest possible value of the sum of their squares is 128.
