Let the two positive numbers be a and b.

Given: a + b = 64 … 1

Also, a³ + b³ is minima

Assume, S = a³ + b³

(from equation 1)

S = a³ + (64 – a)³

= 3a² + 3(64 – a)² × ( - 1)

(condition for maxima and minima)

⇒ 3a² + 3(64 – a)² × ( - 1) = 0

⇒ 3a² + 3(4096 + a² – 128a) × ( - 1) = 0

⇒ 3a² – 3 × 4096 - 3a² + 424a = 0

⇒ a = 32

Since, > 0 ⇒ a= 32 will give minimum value

Hence, two numbers will be 32 and 32.