A manufacturer has been selling 1000 flat-screen TVs a week at $350 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of TVs sold will increase by 100 per week. (a) Find the demand function (price p as a function of units sold x). p(x) = Correct: Your answer is correct. (b) How large a rebate should the company offer the buyer in order to maximize its revenue? $ 125 Correct: Your answer is correct. (c) If its weekly cost function is C(x) = 61,000 + 110x, how should the manufacturer set the size of the rebate in order to maximize its profit?

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

$2.00 for three shots.

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