Answer:
A. p(x) = (-1/10)x + 450x
B. 506,250
C. $280
Explanation:
(a) Calculation to Find the demand function
Let assume that the demand Q(p) is the LINEAR function of the price
Q(350) = 1000
dQ/dp = -100/10 = -10
Point-slope form :
Q-1000 = -10(p-350)
Q-1000 = -10p + 3500
Q= -10p+(3500+1000)
Q = -10p + 4500
Hence,
Q(p) = -10p +4500
The inverse is:
(Q-4500)/-10 = p
(-1/10)Q + 450 = p
Hence, p(x) = (-1/10)x + 450 where Q=x
Therefore the the demand function will be p(x) = (-1/10)x + 450x
(b) Calculation for How large a rebate should the company offer the buyer in order to maximize its revenue
Revenue = Price * Qauntity sold
R(p) = p*Q(p)
Revenue = p * (-10p + 4500)
Revenue= -10p^2 + 4500p
let Maximize:
0 = dR/dp = -20p + 4500
-4500 = -20p
p=-4500/-20
p = 450/2 = 225
Hence, Max. revenue
-10 (225)^2 + 4500 *225
= -506,250 + 1,012,500
= 506,250
Therefore How large a rebate should the company offer the buyer in order to maximize its revenue will be 506,250
(c) Calculation for how should the manufacturer set the size of the rebate in order to maximize its profit
Let C(x) represent cost to produce x television sets
Let C(Q(p)) represent cost to produce the demanded quantity
C(q(p)) = 61,000 + 110*Q(p)
C(q(p)) = 61,000 + 110* (-10p +4500)
C(q(p)) = 61,000 + -1,100p + 495,000
C(q(p)) = 556,000 - 1,100p
let calculate the profit using this formula
Profit = Revenue - Cost
Let plug in the formula
P(p) = -10p^2 + 4,500p - ( 556,000 - 1,100p)
P(p) = -10p^2 + 4,500p -556,000 + 1,100p
P(p) = -10p^2 + 5,600p - 556,000
Maximizing:
0=dP/dp = -20p + 5,600
Hence, maximizing profit occurs at price p=-5,600/-20 = $280
Therefore how should the manufacturer set the size of the rebate in order to maximize its profit will be $280
