A manufacture has been selling 1400 television sets a week at $480 each. A market survey indicates that for each $24 rebate offered to a buyer, the number of sets sold will increase by 240 per week.
a) Find the price p(x) of each television as a function of x, where x is the number of the television sets sold per week.
p(x)=
b) How large a rebate should the company offer, in order to maximize its revenue? dollars
c) If the weekly cost function is 112000+160x, what value of the rebate maximizes the profit? dollars
a) To find the price function, we need to determine how the price changes as the number of television sets sold per week changes. We know that initially, the manufacturer is selling 1400 television sets a week at $480 each. We also know that for each $24 rebate offered, the number of sets sold increases by 240 per week.
Let's assume x represents the number of television sets sold per week. As the number of sets sold increases, the price will be reduced by a certain amount due to the rebate.
Initially, the price is $480. For every 240 additional sets sold, the price decreases by $24. We can express this relationship mathematically:
Price decrease = (Number of additional sets sold / 240) * $24
Therefore, the price function p(x) can be determined as follows:
p(x) = $480 - [(x - 1400)/240] * $24
b) To maximize revenue, we need to determine the price at which the most number of sets will be sold. Since we have the equation for p(x), we can differentiate it with respect to x and set the derivative equal to zero.
dp(x)/dx = -24/240 = -0.1
To maximize revenue, we set dp(x)/dx = 0 and solve for x:
-0.1 = 0
x = 1400
Therefore, the company should offer a rebate that results in 1400 television sets sold per week in order to maximize its revenue.
c) To find the value of the rebate that maximizes profit, we need to consider the cost function. The profit function can be determined as follows:
Profit = Revenue - Cost
Given the cost function of 112000 + 160x, and knowing that revenue is calculated by multiplying the price obtained in part (b) by the number of sets sold, we can express profit mathematically:
Profit = (p(x) * x) - (112000 + 160x)
Substituting the value of x that maximizes revenue (1400) into p(x) and simplifying, we have:
Profit = (p(1400) * 1400) - (112000 + 160 * 1400)
Profit = ($480 - [(1400 - 1400)/240] * $24) * 1400 - (112000 + 160 * 1400)
Now, we can calculate the profit for different rebate values and find the rebate that maximizes it. We'll start by calculating the profit for a $0 rebate:
Profit = ($480 - [(1400 - 1400)/240] * $24) * 1400 - (112000 + 160 * 1400) = $672000
Next, let's calculate the profit for a $24 rebate:
Profit = ($480 - [(1400 - 1400)/240] * $24) * 1400 - (112000 + 160 * 1400 + 240 * (1400 - 1400)) = $683200
Continuing this process, we can calculate the profit for different rebate values and observe which value maximizes it.