Q. 25.0( 2 Votes )

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

Answer :

Let the two positive numbers be a and b.

**Given:** a + b = 64 … 1

Also, a^{3} + b^{3} is minima

Assume, S = a^{3} + b^{3}

(from equation 1)

S = a^{3} + (64 – a)^{3}

= 3a^{2} + 3(64 – a)^{2} × ( - 1)

(condition for maxima and minima)

⇒ 3a^{2} + 3(64 – a)^{2} × ( - 1) = 0

⇒ 3a^{2} + 3(4096 + a^{2} – 128a) × ( - 1) = 0

⇒ 3a^{2} – 3 × 4096 - 3a^{2} + 424a = 0

⇒ a = 32

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

Hence, two numbers will be 32 and 32.

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