Two positive numbers have a sum of 16. The least possible value of the sum of their squares is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 128
Let the numbers be x and 16 − x, with 0 < x < 16. The sum of their squares is
\( S(x) = x^{2} + (16 - x)^{2} \)
Setting the derivative equal to zero:
\( S'(x) = 2x - 2(16 - x) = 0 \)
\( x = 8 \)
Since \(S''(x) = 4 \gt 0\), x = 8 gives the minimum. Both numbers are 8, and
\( S(8) = 64 + 64 = 128 \)