If `x+y=8`, then maximum valueof `x^(2)y` is

Options

A. `2048/9`
B. `2048/81`
C. `2048/3`
D. `2048/27`

Answer

D

Explanation

Let `A=x^(2)y=x^(2)(8-x)` [ given `x+y=8`]
`impliesA=8x^(2)-x^(3)`
`:.(dA)/(dx)=16x-3x^(2)`
For maxima or minima put `(dA)/(dx)=0`
`implies16x-3x^(2)=0`
`implies x=0, 16/3`
Now, `(d^(2)A)/(dx^(2))=16-6x`
At `x=16/3`
`((d^(2)A)/(dx^(2)))_(x=16/3)=16-32-16lt0`, maxima
`:.` Maximum value `=8(16/3)^(2)-(16/3)^(3)=2048/27`