What is the product of two parts of 20 such that the product of one part and the cube of the other is maximum?

Options

A. 75
B. 91
C. 84
D. 96

Answer

A

Explanation

Let 20 be divided in two parts such that first part =x
`therefore` second part =20-x
Now assume that
`p=x^(3)-x^(4)`
Now `(dp)/(dx)=60x^(2)-4x^(3)`
and `(d^(2)p)/(dx^(2))=120x-12x^(2)`
Put `(dp)/(dx)=0` for maxima or minima
`rarr (dp)/(dx)=0`
`rarr 4x^(2)(15-x)=0 rarr x=0 x =15`
`therefore` P sia maximum x =15
and second part =20 - 15=5
Required product `=15xx5=75`