Let x and y be the two numbers (ignoring the fact that x and y are odd for now).

We know that x+y=60. We want to maximize the value of xy.

xy\ =\ x\left(60\ -x\right)\ =\ 60x\ -\ x^2.

If you take a function and differentiate the function, the differential will be 0 at two places - the maxima and the minima. This is true because at the maxima and minima, the function changes direction - i.e., the slope changes from being +ve to -ve passing through 0 at the maxima and minima.

\frac{d\left(60x\ -\ x^2\right)}{dx}\ =\ 60\ -\ 2x

The maxima/minima are at x = 30.

At x = 30, 60x\ -\ x^2 = 1800 - 900 = 900.

At x = 1, 60x\ -\ x^2 = 59.

Hence, at x = 30, we have a maxima.

Given that the two numbers have to be odd, the closest values to 30 are 29 and 31.