MAXIMIZING REVENUE WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

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  =  (y₂ - y₁) / (x₂ - x₁)

Slope  =  (100 - 150) / (45 - 40)

Slope  =  -50 / 5

Slope  =  -10

Price demand equation :

(y - y₁)  =  m(x - x₁)

(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)  =  -10x² + 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

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 :

(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 - y₁)  =  m(x - x₁)

(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)  =  -0.0025x² + 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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