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

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

Given: $x + y = 60$x+y=60

Therefore: $y = 60 - x$y=60−x

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To minimize the sum of squares $x^2 + y^2$x2+y2, substitute $y = 60 - x$y=60−x:

$$x^2 + (60 - x)^2$$x2+(60−x)2

$$= x^2 + 3600 - 120x + x^2$$=x2+3600−120x+x2

$$= 2x^2 - 120x + 3600$$=2x2−120x+3600

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Taking the derivative with respect to $x$x:

$$\frac{d}{dx}(2x^2 - 120x + 3600) = 4x - 120$$dxd​(2x2−120x+3600)=4x−120

Setting equal to zero:

$$4x - 120 = 0$$4x−120=0

$$4x = 120$$4x=120

$$x = 30$$x=30

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Since $x = 30$x=30:

$$y = 60 - 30 = 30$$y=60−30=30

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The absolute value of the difference of their cubes:

$$|x^3 - y^3|$$∣x3−y3∣

$$= |30^3 - 30^3|$$=∣303−303∣

$$= |27000 - 27000|$$=∣27000−27000∣

$$= 0$$=0

Therefore, the absolute value of the difference of their cubes is $0$0.