A national distributor of pet toys determines the cost and revenue of functions for one of its toys.

Question

C( x )= 1.2x−.0001x^2 0≤ x ≤6000

R( x )= 3.6x−.0005x^2 0≤ x ≤6000

Determine the open interval on which the profit function is increasing. Find any extrema.

Answer 1

Profit = P(x) = R(x)-C(x) = 3.6x -.0005x^2 - 1.2x +.0001x^2

= 2.4x - .0006x^2

P'(x) = 2.4 -.0012x =0

x = 2.4/.0012 = 2000 when Profit = max at 2,000

P''(x) = -.0012x is always negative for 0<x<6,000. the graph is always concave down

2nd derivative is negative means 1st derivative is decreasing

but profit is always increasing as output increases from 0 to 2,000

then it decreases from 2,000 to 6,000,

from 4,000 to 6,000, profit is decreasing, but negative, with losses, not profits

try graphing the profit function

It's a downward opening parabola

vertex = maximum profit when x =2,000

after 2,000 profit decreases

max profit = $2400

= global and local or relative max or extremum

global minimum or global extremum

is

2.4(6000) -.0006(6000)^2

= a loss of $7200 or negative $7200 "profit"