Question

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?

Answer 1

To find the demand function, we relate the initial sales information to the rebate and quantity sold. For maximizing revenue, we calculate the derivative of the revenue function to find the optimal number of units sold and corresponding rebate. To maximize profit, we subtract the cost function from the revenue function and find where the derivative is zero.

Step-by-step explanation:

Finding the Demand Function

To find the demand function p(x), we know that at $350 each, 1000 TVs are sold. For each $10 rebate, an additional 100 TVs are sold. The rebate decreases the price, so let the number of $10 rebates be r. Then, the price can be expressed as p(x) = 350 - 10r. Also, for each rebate offered, the quantity sold increases by 100, so the relation between r and the quantity sold x is x = 1000 + 100r. Solving for r in terms of x, we get r = (x - 1000)/100. Substituting r back into the price function gives p(x) = 350 - 10((x - 1000) / 100).

Maximizing Revenue

The manufacturer's revenue R is the product of the number sold x and the price p(x), R(x) = x * p(x). To maximize revenue, find the derivative of the revenue function, set it to 0, and solve for x. The corresponding rebate can then be found using the decrease in price at the optimal number of units sold.

Maximizing Profit

Profit is defined as revenue minus cost. With the cost function C(x) = 61000 + 110x, the profit function is P(x) = R(x) - C(x). Find the derivative of P(x), set it to 0, and solve for x to find the quantity that maximizes profit. The corresponding rebate size will be computed based on the optimal number of TVs sold. The rebate amount is critical to maximizing profit, as it affects both the quantity sold and the margin per unit.

Answer 2

A. p(x) = (-1/10)x + 450x

B. 506,250

C. $280

Step-by-step 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