MHT CET202520 Apr 2025Morning ShiftMathematicsApplication of DerivativesActual
20 is divided into two parts so that the product of the cube of one part and the square of the other part is maximum, then these two parts are
Options
- A15,5
- B16,4
- C12,8
- D14,6
Correct answer
C. 12,8
Step-by-step solution
Let the two parts be x and y such that x + y = 20 and P = x^3 y^2 . Expressing y = 20 - x gives P(x) = x^3 (20 - x)^2 . Differentiate using the product rule: P'(x) = 3x^2 (20 - x)^2 - 2x^3 (20 - x) = x^2(20 - x)(60 - 5x) . Set P'(x) = 0 to find critical points: x = 0 , x = 20 , or x = 12 . The endpoints x = 0 and x = 20 yield P = 0 , while x = 12 gives y = 8 and P = 12^3 8^2 = 110592 . The derivative changes from positive to negative at x = 12 , confirming a maximum. Among the options, C corresponds to 12 and 8 wit