The price p and the quantity x sold of a small flatscreen te

The price p and the quantity x sold of a small flat-screen television set obeys the demand equation below. a) How much should be charged for the television set if there are 80 television sets in stock? b) What quantity x will maximize revenue? What is the maximum revenue? c) What price should be charged in order to maximize revenue? p = 20x + 580 a) What should be charged for the units in stock? exist b) What quantity will maximize the revenue? television sets What is the maximum revenue? exist c) What price should be charged for the maximum revenue? exist

Solution

a) If there are 80 TVs in stock, and all are to be sold, then x = 80 so that p = -0.2*80 +580 = -16+580 = 564. Thus, 564 should be the price to be charged if all t5he 80 TVs are to be sold.

b) The revenue function R(X) = price* quantity sold = px = (-0.2x +580)x = -0.2x²+580x. Now, R(x) is maximum when dR/dx= 0 and d²R/dx² is negative. Here, dR/dx = -0.4x+580 so that dR/dx = 0 when 0.4x = 580 or, x = 580/0.4 = 1450. Further, d²R/dx²= -0.4 is always negative regardless of the value of x. Thus, x = 1450 will maximize the revenue. Also, when x = 1450, we have p = -0.2*1450 +580 = -290+580 = 290. Thus, the maximum revenue is 1450*290= 420500.

c) The price to be charged is 290, if the revenue is to be maximized.