Let the two numbers are x and y such that x + y = 60

⇒ y = 60 –x

Let f(x) = xy³

⇒ f(x) = x(60-x)³

⇒ f’(x) = (60 –x)³ -3x(60 –x)²

= (60-x)²[60 – x – 3x]

= (60-x)²[60 – 4x]

And f’’(x) = -2(60 – x)(60 -4x) -4(60-x)²

= -2(60 – x)[60 -4x + 2(60-x)]

= -2(60 – x)(180 – 6x)

= -12(60 – x)(30 -x)

Now, f’(x) =0

⇒ x = 60 or x =15

When x = 60, f’’(x) = 0.

When x = 15, f’’(x) = -12(60-15)(30-15) = -12×45×15 < 0

Then, by second derivative test, x =15 is a point of local maxima of f.

Then, function xy³ is maximum when x =15 and y = 60 – 15 = 45.

Therefore, required numbers are 15 and 45.