**Problem 1 :**

A company has determined that if the price of an item is $40, then 150 will be demanded by consumers. When the price is $45, then 100 items are demanded by consumers.

(a) Find the price-demand equation, assuming that it is linear.

(b) Find the revenue function.

(c) Find the number of items sold that will give the maximum revenue. What is the maximum revenue ?

(d) What is the price of each item when maximum revenue is achieved ?

**Solution :**

Let x be the price and y be the demand.

Price = independent variable and demand = dependent variable. Since the relationship between price and demand is linear, we can form a equation.

(40, 150) and (45, 100)

Slope = (y_{2} - y_{1}) / (x_{2} - x_{1})

Slope = (100 - 150) / (45 - 40)

Slope = -50 / 5

Slope = -10

Price demand equation :

(y - y_{1}) = m(x - x_{1})

(y - 150) = -10(x - 40)

y - 150 = -10x + 400

y = -10x + 400 + 150

y = -10x + 550

Price of item per unit.

(b) Revenue function

Revenue = Units Sold x Sales Price

= (-10x + 550) ⋅ x

R(x) = -10x^{2} + 550x

(c) To find the number of units sold to get the maximum revenue, we should find "y" coordinate at the maximum point.

x coordinate at maximum = -b/2a

x = 550/20 x = 55/2 x = 27.5 |
y = -10 (27.5) + 550 y = -275 + 550 y = 275 |

when 275 units sold, we can get the maximum revenue.

(d)

R(27.5) = -10(27.5)^{2} + 550(27.5)

= -7562.5 + 15125

= $7562.5

The maximum revenue is $7562.5.

**Problem 2 :**

A deli sells 640 sandwiches per day at a price of $8 each. A market survey shows that for every $0.10 reduction in price, 40 more sandwiches will be sold.

(a) Find the linear price-demand function.

(b) Find the revenue equation.

(c) How many sandwiches should be sold to maximize the revenue ?

(d) How much should the deli charge for a sandwich in order to maximize its revenue ?

**Solution :**

Number of sandwiches 640 680 720 |
Cost per sandwich $8 $7.9 $7.8 |

(a) Let x the number of sandwiches and y be the cost per sandwich.

Price demand function :

(640, 8) and (680, 7.9)

m = (7.9 - 8)/(680 - 640)

m = -0.1/40

m = - 0.0025

(y - y_{1}) = m(x - x_{1})

(y - 8) = -0.0025(x - 640)

y = -0.0025x + 1.6 + 8

y = -0.0025x + 9.6

(b)

Revenue = Units Sold x Sales Price

= (-0.0025x + 9.6) x

R(x) = -0.0025x^{2} + 9.6x

(c) We should find the number of sandwiches to be sold out to maximize the revenue.

x = -b/2a

x = 9.6/2(0.0025)

x = 1920 (number of sandwiches)

To get the maximum revenue, 1920 sandwiches to be sold out.

(d) Cost per sandwich

y = -0.0025x + 9.6

x = 1920

y = -0.0025(1920) + 9.6

y = -4.8 + 9.6

y = 4.8

Deli has to charge $4.8 for a sandwich in order to maximize its revenue.

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