Question

1) The relationship d = 5000 - 25p describes what happens to demand (d) as price (p) varies. Here, price can vary between $10 and $50.

a) How many units can be sold at the $10 price? How many can be sold at the $50 price?

b) Model the expression for total revenue.

c) Consider prices of $20, $30, and $40. Which price alternative will maximize total revenue? What are the values for demand and revenue at this price?

Answer 1

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a. At $10 demand will be 5000 - (25*10) = 5000 - 250 = 4750 units

At $50 demand will be = 5000-(25*50) = 5000 - 1250 = 3750 units

b. Total revenue = total demand (quantity in units)*price (in $)

d = 5000-25p and price = p. Thus total revenue = (5000-25p)*p

= 5000p-25p^2

c. Total revenue and demands at different price points:

$20: demand = 5000-(25*20) = 5000-500 = 4500

revenue = 5000*20 - (25*20*20) = 90,000

$30: demand = 5000-(25*30) = 4250

revenue = 5000*30 - (25*30*30) = 127500

$40: demand = 5000-(25*40) = 4000

revenue = 5000*40 - (25*40*40) = 160000

Thus total revenue is maximized at $40 and the amount of revenue is $160,000

The demand at this price = 4,000 units and revenue = $160,000

ALTERNATIVELY:

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