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
- A(12,8)
- B(15,5)
- C(10,10)
- 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) .