The profit fucntion in rupees of a firm selling x items `(x ge 0)` per week is given by `P(x) =-3500 +(400-x)x` How many items should the firm sell so that the firm has maximum profit ?

Options

A. 400
B. 300
C. 200
D. 100

Answer

C

Explanation

`p(x)=-3500 +(400-x)x=-3500+400x - x^(2)`
pone differentiationg w.r.t x we get
p(x) =400 -2x
Put p'(x) =0 for maxima or minima
`rarr 400-2x=0`
`rarr x=200`
Now `p''(x)=-2x`
`rarr P''(200)=-400 lt 0`
`therefore` p(x) is maximum at x=200
Hence 200 items should the firm sell so that the firm has maximum profit