The sum of two positive numbers is 60. If the sum of their squares is…

The correct answer is Option (1) → 0

Let the numbers be \(x\) and \(60 - x\), where \(0 \lt x \lt 60\). The sum of their squares is

    \(S = x^2 + (60 - x)^2\)

    \( \displaystyle \frac{dS}{dx} = 2x - 2(60 - x) = 4x - 120 \)

Setting \(\frac{dS}{dx} = 0\):

    \(4x - 120 = 0,\quad x = 30\)

    \( \displaystyle \frac{d^2S}{dx^2} = 4 \gt 0 \)

So by the second derivative test, S is minimum at \(x = 30\), and the other number is also \(60 - 30 = 30\).

    \(|30^3 - 30^3| = 0\)

Shortcut: \(x^2 + y^2 = \frac{1}{2}\left[(x + y)^2 + (x - y)^2\right] = 1800 + \frac{1}{2}(x - y)^2\), which is least when \(x = y\). Equal numbers have cubes with difference 0.