Q. 25.0( 2 Votes )
Divide 64 into two parts such that the sum of the cubes of two parts is minimum.
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
(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.
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