Question

Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Answer 1

Suppose the two positive numbers be a and b.

Given as a + b = 64 … (1)

We have, a³ + b³ is minima

Assume that, S = a³ + b³

(From the equation 1)

S = a³ + (64 – a)³

Thus, (d²S/da²) > 0 

a = 32 will give the minimum value 

Thus, the two number will be 32 and 32