The sum of three positive numbers is 26. The second number is thrice as large as the first. If the sum of the squares of these numbers is least, find the numbers.

4, 12, 10

The correct answer is Option (2) → 4, 12, 10

Let the first number be $x (0 < x < 26)$, then the second number is $3x$. Since the sum of the three numbers is 26, therefore, the third number is $26-x-3x$ i.e. $26-4x$.

Let $f(x) = x^2+ (3x)^2 + (26 - 4x)^2 = 26x^2 - 208x + 676$   ...(i)

Differentiating (i) w.r.t. x, we get

$f'(x) = 52x - 208$ and

$f''(x) = 52$.

Now $f'(x) = 0 ⇒ 52x - 208 = 0⇒x= 4$.

Also $f''(4) = 52 > 0$ $⇒f(x)$ is minimum at $x = 4$.

Hence, the required numbers are 4, 3 × 4 and 26 - 4 × 4 i.e. 4, 12 and 10.