Divide 64 into two parts such that the sum of the cubes of two parts i

1.	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, a³ + b³ is minima Assume, S = a³ + b³ (from equation 1) S = a³ + (64 – a)³ ⇒ \(\frac{ds}{da}\) = 3a² + 3(64 – a)² × ( - 1) ⇒ \(\frac{ds}{da}\) = 0 (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 \(\frac{d^2s}{da^2}\) = 6a + 6(64 - a) = 384 Since, \(\frac{d^2s}{da^2}\)> 0 ⇒ a = 32 will give minimum value Hence, two numbers will be 32 and 32.

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,

a³ + b³ is minima

Assume,

S = a³ + b³ (from equation 1)

S = a³ + (64 – a)³

⇒ \(\frac{ds}{da}\) = 3a² + 3(64 – a)² × ( - 1)

⇒ \(\frac{ds}{da}\) = 0

(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

\(\frac{d^2s}{da^2}\) = 6a + 6(64 - a) = 384

Since,

\(\frac{d^2s}{da^2}\)> 0

⇒ a = 32 will give minimum value

Hence, two numbers will be 32 and 32.