Raymond B. answered 06/19/21
(195, 310) (190, 320) are two points on a graph of (Q,P) or (x,y) with price measured on the y axis and number of suites rented on the x axis
line through the points has slope = (320-310)/(190-195) = 10/-5 = -2
(y-320)/(x-190) = -2
-2x + 380 = y-320
y = -2x +700, at a price of $700, no rooms are rented
at a price of $350, 175 rooms are sold and revenue = 175(350)= $61,250
at $300 revenue = 300(200) = $60,000, less than at a price of $350
revenue R = PQ = yx = -2x^2 + 700x
R' = -4x + 700 = 0
x = 175 = revenue maximizing quantity
y=-2(175)+700 = $350 = revenue maximizing price
R= xy= PQ = 175(350) = $61,250 = maximum revenue