Divide 20 into two parts such that the product of the cube of one and the square of the other shall be maximum

Options

A. 6,14
B. 12,8
C. 10,10
D. 5,15

Answer

B

Explanation

`f(x)=x^(3)(20-x)^(2)`
`:.f(x)=x^(3)(400-40x+x^(2))=x^(5)-40x^(4)+400x^(3)`
`impliesf'(x)=5x^(4)-160x^(3)+1200x^(2)`
`=5x^(2)(x^(2)-32x+240)`
For maximum put `f'(x)=0`
therefore `x^(2)-32x+240=0`
i.e. `(x-12)(x-20)=0`
`:.x=12` or `x!=20`
`:.20-x=20-12=8`
Now `f''(x)=20x^(3)-480x^(2)+2400x`
`impliesf''(12)=34560-69120+28800=-5760lt0`
Here `f` is maximum at `x=12`
Hene 12 and 8 are two part of 20.