Question

A manufacturer has been selling 1000 television sets a week at $450 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 demand function (price as a function of units sold). p(x) = (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) = 74,000 + 150x, how should the manufacturer set the size of the rebate in order to maximize its profit?

Answer 1

Answer:

a) P(x) = 550 - x/10

b) $175

c) $100

Step-by-step explanation:

Number of television sold per week = 1000

The selling price for each = $450

a) If $10 rebate offered to the buyer increases the sale by 100 per week, the price will decrease by

1/100*10 = 1/10 per unit

Let X be the number of television sold per week

The increase in number sold per week = x - 1000

P(x) = 450 - 1/10(x-1000)

P(x) = 450 - x/10 + 100

= 550 - x/10

The demand function is

P(x) = 550 - x/10

b) Let R(x) be the revenue

R(x) = xP(x)

= x[550 - x/10]

= 550x - x^2/10

Differentiate R(x) with respect to x

R'(x) = 550 - 2x/10

= 550 - x/5

R'(x) is the marginal revenue

The revenue is maximized when R'(x) = 0

550 - x/5 = 0

550 = x/5

x = 550*5

x = 2750

Recall that P(x) = 550 - x/10

Put x= 2750

P(2750) = 550 - 2750/10

= 550 - 275

= 275

The rebate to maximize the revenue will be 450 - 275 = $175

c) C(x) = 74000 + 150x

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

P(x) = 550x - x^2/10 - (74000 + 150x)

= 550x - x^2/10 - 74000 - 150x

Collect like terms

P(x) = 550x - 150x - x^2/10 - 74000

= 400x - x^2/10 - 74000

Differentiate P(x) with respect to x

P'(x) = 400 - 2x/10

= 400 - x/5

P'(x) is the marginal profit.

The profit is maximum when P'(x) = 0

400 - x/5 = 0

400 = x/5

x = 400*5

x = 2000

Recall that P(x) = 550 - x/10

P(2000) = 550 - 2000/10

= 550 - 200

= 350

The rebate to maximize the revenue = 450 - 350

= $100