# A manufacture has been selling 1050 television sets a week at \$480 each. A market survey indicates that for each \$25 rebate offered to a buyer, the number of sets sold will increase by 250 per week.

a) Find the function representing the demand p(x), where x is the number of the television sets sold per week and p(x) is the corresponding price.
p(x)=

b) How large rebate should the company offer to a buyer, in order to maximize its revenue? dollars

c) If the weekly cost function is 84000+160x, how should it set the size of the rebate to maximize its profit? dollars

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## a) The demand function, p(x), can be found by analyzing the given information. Initially, the company sells 1050 television sets a week at \$480 each. As the number of sets sold increases by 250 for every \$25 rebate offered, we can represent the demand function as follows:

p(x) = (1050 + 250x) * (480 - 25x)

b) To maximize revenue, we need to find the value of x that maximizes p(x). We can do this by finding the maximum point of the demand function. This can be done by taking the derivative of p(x) and setting it equal to zero:

p'(x) = 250(480 - 25x) - 25(1050 + 250x) = 0

Solving for x:

250(480 - 25x) - 25(1050 + 250x) = 0
120000 - 6250x - 26250 - 6250x = 0
185000 - 12500x = 0
-12500x = -185000
x = 14.8

Since we can't sell fractional television sets, we can round x down to the nearest whole number, giving us x = 14.

Now, substitute x = 14 into the demand function to find the corresponding price:

p(14) = (1050 + 250*14) * (480 - 25*14)
p(14) = (1050 + 3500) * (480 - 350)
p(14) ≈ 4550 * 130
p(14) ≈ 591,500

Therefore, to maximize revenue, the company should offer a \$14 rebate to the buyers.

c) To maximize profit, we need to consider the cost function as well. The profit function can be calculated by subtracting the cost from the revenue:

profit(x) = revenue(x) - cost(x)
profit(x) = x * p(x) - (84000 + 160x)

Now, set the derivative of the profit function equal to zero to find the value of x that maximizes profit:

profit'(x) = p(x) + xp'(x) - 160 = 0

Substitute x = 14 into the profit function to find the corresponding profit:

profit(14) = 14 * p(14) - (84000 + 160*14)
profit(14) ≈ 14 * 591,500 - 84000 - 2240
profit(14) ≈ 8,280,500 - 84000 - 2240
profit(14) ≈ 8,195,260

Therefore, to maximize profit, the company should offer a \$14 rebate to the buyers. The profit will be approximately \$8,195,260.

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## a) To find the demand function, we need to determine the relationship between the price (p) and the number of sets sold per week (x).

Given that the manufacturer has been selling 1050 television sets a week at a price of \$480 each, we can start by finding the initial demand equation based on this data.

We know that when the price is \$480, the number of sets sold is 1050. This gives us the point (1050, 480) on the demand curve.

Let's assume that for each \$25 rebate offered to a buyer, the number of sets sold will increase by 250 per week. This means that if we decrease the price by \$25 (from \$480 to \$455), the number of sets sold will increase by 250.

Using this information, we can define the demand function as follows:

p(x) = (1050 + 250(x - 1050/250)) * 25

Let's simplify this equation step by step:

p(x) = (1050 + 250(x - 4.2)) * 25 [simplifying 1050/250 = 4.2]

p(x) = (1050 + 250x - 1050) * 25 [distributing 250(x - 4.2)]

p(x) = 250x * 25

p(x) = 6250x

Therefore, the demand function is p(x) = 6250x.

b) To find the rebate amount that maximizes revenue, we need to consider the revenue function.

Revenue (R) can be calculated as the product of the price (p) and the quantity sold (x). In this case, the revenue function is:

R(x) = p(x) * x

Substituting the demand function p(x) = 6250x, we get:

R(x) = 6250x * x

R(x) = 6250x^2

To find the value of x that maximizes revenue, we can take the derivative of the revenue function with respect to x and set it equal to zero:

R'(x) = 12500x = 0

Solving for x:

12500x = 0
x = 0

Since x = 0 is not a valid solution (as we cannot sell zero sets), we can conclude that the revenue is maximized at the critical point where the derivative equals zero.

Therefore, the company should offer a rebate amount that maximizes its revenue is \$0 (no rebate at all).

c) To maximize profit, we need to consider the cost function in addition to the revenue function.

The profit function (P) is given by subtracting the cost function (C) from the revenue function (R):

P(x) = R(x) - C(x)

Since we already have the revenue function R(x) = 6250x^2, let's substitute the given cost function C(x) = 84000 + 160x:

P(x) = 6250x^2 - (84000 + 160x)

Expanding and simplifying the expression:

P(x) = 6250x^2 - 84000 - 160x

To find the value of x that maximizes profit, we can take the derivative of the profit function with respect to x and set it equal to zero:

P'(x) = 12500x - 160 = 0

Solving for x:

12500x = 160
x = 160/12500

Calculating the value of x:

x ≈ 0.0128

Therefore, to maximize its profit, the company should set the size of the rebate to approximately \$0.0128 (or around 1.28 cents).