The maximum value of `xy` when `x+2y=8` is

Options

A. 20
B. 16
C. 24
D. 8

Answer

D

Explanation

Given `x+2y=8impliesy=(8-x)/2`……..i
let `p=xy=(8x-x^(2))/2implies(dp)/(dx)=1/2(8-2x)`
For maxima or minima, put `(dp)/(dx)=0`
`implies8-2x=0impliesx=4`
Again `(d^(2)p)/(dx^(2))=-1`
At `x=4`, ((d^(2)p)/(dx^(2)))=-1lt0`
At `x=4, ((d^(2)p)/(dx^(2)))_(x=4)=-1lt0`
From Eq. i `y=2`
Thus, function is maximum at `x=4` and `y=2`
Therefore maximum value of `p=4xx2=8`