A manufacturer has been selling 1000 television sets a week at $500 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. Round your answers to the nearest dollar.

(a) Find the linear demand function (price as a function of units sold).

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

(c) If the company experiences a cost of C (x) = 76,000 + 110x, how should the manufacturer set the size of the rebate in order to maximize its profit?

Question

Mathematics

Silas Gonzales

26 October, 18:03

A manufacturer has been selling 1000 television sets a week at $500 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. Round your answers to the nearest dollar.

(a) Find the linear demand function (price as a function of units sold).

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

(c) If the company experiences a cost of C (x) = 76,000 + 110x, how should the manufacturer set the size of the rebate in order to maximize its profit?

+3

Answers (1)

Know the Answer?

Abdiel Foster · Answer 1

a) The linear function is

P (x) = 600 - x/10

b) $200

c) $145

Step-by-step explanation:

The manufacturer sold a total number of 1000 television in a week

The selling price of each television is $500

The sale increases by 100 per week if a rebate of $10 dollars is offered to the buyer.

Therefore, we have a price decrease of 1/100 * 10 = 1/10 for each unit

Let x be the number sold per week.

x-1000 gives the increase in sales

Let P (x) be the price

P (x) = 500 - 1/10 (x - 1000)

= 500 - x/10 + 1000/10

= 500 - x/10 + 100

Collect like terms

P (x) = 500 + 100 - x/10

= 600 - x/10

The linear demand function with price as a function of units sold is given as

P (x) = 600 - x/10

b) Let R (x) be the revenue

Revenue = number sold * Price

R (x) = x * P (x)

R (x) = x (600 - x/10)

= 600x - x^2/10

Differentiate R (x) with respect to y

R' (x) = 600 - 2x/10

= 600 - x/5

The revenue is maximum when R (x) = 0

600 - x/5 = 0

600 = x/5

x = 600*5

x = 3000

Remember that P (x) = 600 - x/10

P (3000) = 600 - 3000/10

= 600 - 300

= 300

The rebate to maximize the revenue will be 500 - 300 = $200

c) C (x) = 76000 + 110x

C (x) = Cost

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

P (x) = (600x - x^2/10) - (76000 + 110x)

= 600x - x^2/10 - 76000 - 110x

Collect like terms

P (x) = 600x - 110x - x^2/10 - 76000

= 490x - x^2/10 - 76000

Differentiate P (x) with respect to x

P' (x) = 490 - 2x/10

= 490 - x/5

P (x) is maximized when it is equal to 0

490 - x/5 = 0

490 = x/5

x = 490*5

x = 2450

P (x) = 600 - x/10

P (2450) = 600 - 2450/10

= 600 - 245

= 355

The rebate to maximize the revenue will be 500 - 355 = $145