A manufacturer has been selling 1400 television sets a week at $450 each. A market survey indicates that for each $21 rebate offered to a buyer, the number of sets sold will increase by 210 per week. (a) Find the function representing the demand p(x)p(x), where xx is the number of the television sets sold per week and p(x)p(x) is the corresponding price. p(x)

Correct question is;

A manufacturer has been selling 1400 television sets a week at $450 each. A market survey indicates that for each $21 rebate offered to a buyer, the number of sets sold will increase by 210 per week

A) Find the demand function (price p as a function of units sold x).

p(x)=

(B) How large a rebate should the company offer the buyer in order to maximize its revenue?

$=

A) p(x) = (-1/10)x + 590

B) Rebate = $170

Step-by-step explanation:

A) We are told that p(x) is the demand function and x is the number of TV sets sold per week.

Now, since for each $21 rebate offered, the number of sets sold increases by 210 per week, it means that the slope here of this demand function is; m = -21/210 = -1/10

Now, he has been selling 1400 TV sets a week at $450 each. This means; p(1400) = $450

Thus, the demand function will be;

p(x) - 450 = (-1/10)(x - 1400)

Expanding the RHS;

p(x) - 450 = (-1/10)x + 140

Add 450 to both sides to get;

p(x) = (-1/10)x + 140 + 450

p(x) = (-1/10)x + 590

B) Formula for revenue is;

R = price × quantity sold

Our demand function is p = (-1/10)x + 590

Making x the subject, we have;

x = 5900 - 10p

x is quantity sold.

Thus,

R = p(5900 - 10p)

R = 5900p - 10p²

Maximum price will occur at dR/dP = 0

Thus;

dR/dP = 5900 - 20p

At dR/dP = 0,we have;

20p = 5900

p = 5900/20

p = $280

Thus, rebate = 450 - 280 = $170

(-5,-28)

i graphed to on desmos.