A manufacture has been selling 1300 television sets a week at $420 each. A market survey indicates that for each $16 rebate offered to a buyer, the number of sets sold will increase by 160 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? in dollars

c) If the weekly cost function is 91000+140x, what value of the rebate maximizes the profit? in dollars

This is the same type of question as your previous one dealing with rent increases.

Follow my steps, by defining the number of $16 rebates as n

for c) recall that profit = revenue - cost
so just subtract the cost expression from your expression from a) and repeat the steps.

a) To find the price function p(x), we need to consider how the price is affected by the rebate and the number of sets sold.

Let's start by calculating the increase in sets sold per week based on the rebate. We are told that for every $16 rebate offered, the number of sets sold increases by 160 per week. This means that the slope of the increase is 160/16 = 10.

Since the company is selling 1300 television sets a week at $420 each without any rebate, we can start with this price as the base price. Then, for each $16 rebate, the number of sets sold will increase by 160, so the price of each television should decrease accordingly to maintain the same revenue.

Therefore, the price function p(x) can be expressed as:
p(x) = 420 - 10x

b) To maximize revenue, we need to find the number of sets that will generate the highest revenue. The revenue can be calculated by multiplying the price and the number of sets sold: R(x) = p(x) * x.

In this case, we know that x represents the number of sets sold per week. So, to maximize revenue, we need to find the value of x that will make R(x) as large as possible.

To do this, we can take the derivative of R(x) with respect to x and set it equal to zero to find the critical point(s). However, in this case, let's simplify the problem by using common sense.

We know that offering a rebate increases the number of sets sold, but it also lowers the price per set. So, if we offer a large rebate, more sets will be sold, but at a lower price. On the other hand, if we offer no rebate, the number of sets sold will be lower, but at a higher price.

To maximize revenue, we want to find the sweet spot where the increase in sets sold due to the rebate compensates for the decrease in price, resulting in the highest total revenue.

Based on this reasoning, we can conclude that the company should not offer any rebate in order to maximize its revenue. Therefore, the rebate size should be $0.

c) To find the rebate size that maximizes profit, we need to consider both revenue and cost. The profit function is given by:
Profit(x) = Revenue(x) - Cost(x)

We know the revenue function is R(x) = p(x) * x. And the cost function is provided as 91000 + 140x.

So, the profit function becomes:
Profit(x) = (420 - 10x) * x - (91000 + 140x)

Now, we can find the value of x that maximizes profit by taking the derivative of the profit function with respect to x, setting it equal to zero, and solving for x:

dProfit/dx = 0
(420 - 10x) - 10x = 0
420 - 20x = 0
20x = 420
x = 21

Therefore, the value of x that maximizes profit is 21. To find the value of the rebate that maximizes profit, substitute x = 21 into the price function p(x):

p(21) = 420 - 10 * 21
p(21) = 420 - 210
p(21) = 210

Hence, the value of the rebate that maximizes the profit is $210.