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COMEDK2012MathematicsDifferentiation

Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are

Options

  1. A(12,8)
  2. B(15,5)
  3. C(10,10)
  4. D(2,18)

Correct answer

B. (15,5)

Step-by-step solution

According to question, x+y=20 ...(i) and p=x y³=(20-y) y³ [by Eq. (i)] p=20 y³-y⁴ d p d y =60 y²-4 y³ d² p d y² =120 y-12 y² Now, d p d y =0 60 y²-4 y³=0 60 y²=4 y³ 4 y=60 y=15 x=5 Hence, two parts are (15,5) .

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