Vivekanand Vellanki
1

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.

Mahesh Godavarti
1

Any odd number can be written as 2k + 1 where k is an even integer.


Let the other odd number be 2m + 1 . Then, we have 2k + 1 + 2m + 1 = 60 \implies k + m = 29 .


We want to maximize k \cdot m = k (29 - k) .


Essentially, we want to find maximum point of the function y = k (29 - k) . Note that this is an inverted quadratic function, therefore, it's maximum occurs at its vertex. For y = ax^2 + bx + c , the vertex occurs at x = -\frac{b}{2a} . Therefore, the vertex occurs at k=\frac{29}{2}. Since, in a quadratic function is a monotonically increasing or decreasing function on the either side of the vertex and k is an integer, we can select k to be closest to \frac{29}{2} . Therefore, k is either 14 or 15 which makes m either 15 or 14 .


In either case, the two odd numbers turn out to be 29 and 31 .