The sum of two positive numbers is 60. If the sum of their squares in minimum, then the absolute value of the difference of their cubes is

0

The correct answer is Option (1) → 0

Let the two positive numbers be $x$ and $y$.

Given

$x+y=60$

Sum of squares:

$S=x^{2}+y^{2}$

Write $y=60-x$

$S=x^{2}+(60-x)^{2}$

$S=2x^{2}-120x+3600$

Differentiate $S$:

$\frac{dS}{dx}=4x-120$

For minimum,

$4x-120=0$

$x=30$

So $y=30$

Now compute absolute difference of cubes:

$|x^{3}-y^{3}|=|30^{3}-30^{3}|$

$=|0|=0$