A manufacturer has been selling 1000 flat-screen TVs a week at $400 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).
(b) How large a rebate should the company offer the buyer in order to maximize its revenue?
(c) If its weekly cost function is C(x) = 73,000 + 110x, how should the manufacturer set the size of the rebate in order to maximize its profit?

Explained below.

Step-by-step explanation:

(a)

It is provided that the price function for 1000 TVs is,

p (1000) = 400.

Also provided that if rebate of $10 is given then sale increases by 100 per week.

Let x be the number of unit sold per week then (x − 1000) is the increase in the number  of units sold.

Then the price function is:

Thus, the demand function is, .

(b)

The revenue function is:

Maximize the revenue as follows:

Observe that R'(x) > 0 for 0 ≤ x < 2500 and R'(x) < 0 for x > 2500. Hence, first  derivative test will lead to the conclusion that maximum occurs at x = 2500.

Compute the value p (2500) as follows:

Then the rebate to maximize the revenue should be: $400 - $250 = $150.

(c)

The weekly cost function is,

Compute the profit function as follows:

Compute the marginal profit as follows:

Compute the value of x for P'(x) = 0 as follows:

Observe that P'(x) > 0 for 0 ≤ x < 1950 and P'(x) < 0 for x > 1950. Hence, first

derivative test will lead to the conclusion that maximum occurs at x = 1950.

Compute the value p (1950) as follows:

Then the rebate to maximize the profit should be: $400 - $305 = $95.

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

can you rephrase the question ? it is unclear : -)

step-by-step explanation: