CUET UG 2025 — Mathematics Calculus

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.