The demand function for a certain make of replacement cartri

Solution

p(x) = .01x^2 - .2x + 54 x = demand for a week, in units of 1000 Price is set at $6 Refer to diagram in Wikipedia article on Economic Surplus (link is below) to determine what the integral is supposed to be The lower limit of the definite integral is zero (the leftmost, lowest x value for the area) To find the upper limit of the definite integral (the upper x value), solve for x: 6 = -.01 x^2 - .2x + 54 Multiply through by 100 to make everything integers so we can see it better 600 = -x^2 - 20 x + 5400 -x^2 - 20 x + 4800 = 0 x^2 + 20 x - 4800 = 0 Complete the square (or your alternate favorite quadratic solving method. Personally I hate to switch windows because I lose the tab) x^2 + 20 x + 100 - 100 - 4800 = 0 (x + 10)^2 - 4900 = 0 (x + 10)^2 = 4900 x + 10 = +/- 70 x = +70 - 10 = 60 <- use this one x = -70 - 10 = -80 <- not feasible Integrate for x = 0 to 60 p(x)-6 dx Integral from 0 to 60 -0.1x^2 - .2x + 48 = -0.01/3 x^3 - 0.1 x^2 + 48x evaluated at 60, minus evaluated at 0 (which is zero, so just evaluate at 60) -0.01/3 * 60^3 - 0.1 * 60^2 + 48*60 60 * (48 - 0.1 * 60 - 0.01 * 60^2 / 3) (There are definitely quicker ways to get this computed) 60 * (48 - 6 - 36/3) 60 * (48 - 6 - 12) 60 * 30 = 1800 answer