Math 201-103-RE - Calculus I Business Functions In ...

Math 201-103-RE - Calculus I Business Functions

Application of the derivative (2) Business and Economics

In Business, the following functions are important. Revenue function = (price per unit) . (quantity of units)

Symbols: R = p . x

Cost function = (average cost per unit) . (quantity of units)

Symbols: C = C . x Profit function = revenue - cost Symbols: P = R - C Sometimes in a problem some of these functions are given. Note: Do not confuse p and P . The price per unit p is also called the demand function p .

Page 1 of 15

Marginal Functions: The derivative of a function is called marginal function.

The derivative of the revenue function R(x) is called marginal revenue with notation:

R

(x)

=

dR dx

The derivative of the cost function C(x) is called marginal cost with notation:

C

(x)

=

dC dx

The derivative of the profit function P (x) is called marginal profit with notation:

P

(x)

=

dP dx

Example 1: Given the price in dollar per unit p = -3x2 + 600x , find: (a) the marginal revenue at x = 300 units. Interpret the result.

revenue function: R(x) = p . x = (-3x2 + 600x) . x = -3x3 + 600x2

marginal revenue: R (x) = dR = -9x2 + 1200x dx

marginal revenue at x = 300 = R (300) = dR

= -9(300)2 + 1200(300) = -450 000

dx x=300

Interpretation: If production increases from 300 to 301 units, the revenue decreases by 450 000 dollars.

(b) the marginal revenue at x = 100 units. Interpret the result.

revenue function: R(x) = p . x = (-3x2 + 600x) . x = -3x3 + 600x2

marginal revenue: R (x) = dR = -9x2 + 1200x dx

marginal revenue at x = 100 = R (100) = dR

= -9(100)2 + 1200(100) = 30 000

dx x=100

Interpretation: If production increases from 100 to 101 units, the revenue increases by 30 000 dollars.

Math 201-103-RE - Calculus I

Application of the derivative (2) Business and Economics

Example 2: Given the average cost in dollar per unit C = 357x + 1800 , find:

the marginal cost at x = 50 units. Interpret the result.

cost function: C(x) = C . x = (357x + 1800) . x = 357x2 + 1800x

dC marginal cost: C (x) = dx = 714x + 1800

marginal cost at x = 50 = C (50) = dC

= 714(50) + 1800 = 37 500

dx x=50

Interpretation: If production increases from 50 to 51 units, the cost increases by 37 500 dollars.

Page 2 of 15

Example 3: Given the revenue function in dollars R(x) = -3x3 + 600x2 and the cost function in dollars C(x) = 357x2 + 1800x ; find:

(a) the marginal profit at x = 10 units. Interpret the result.

profit function = revenue - cost

P (x) = (-3x3 + 600x2) - (357x2 + 1800x) = -3x3 + 243x2 - 1800x

marginal profit:

P

(x) =

dP dx

= -9x2 + 486x - 1800

marginal profit at

x = 10

=

P (10) =

dP dx

= -9(10)2 + 486(10) - 1800 = 3300

x=10

Interpretation: If production increases from 10 to 11 units, the profit increases by 3300 dollars.

(b) the marginal profit at x = 100 units. Interpret the result.

profit function = revenue - cost

P (x) = (-3x3 + 600x2) - (357x2 + 1800x) = -3x3 + 243x2 - 1800x

marginal profit: P (x) = dP = -9x2 + 486x - 1800 dx

marginal profit at x = 100 = P (100) = dP

= -9(100)2 + 486(100) - 1800 = -750 000

dx x=100

Interpretation: If production increases from 100 to 101 units, the profit decreases by 750 000 dollars.

Math 201-103-RE - Calculus I

Application of the derivative (2) Business and Economics

Page 3 of 15

Maximum-Minimum Problems: Optimization

To optimize a function means the following: To maximize the revenue function To minimize the cost function To maximize the profit function. Procedure: (a) Define a variable x and build the equation of a function based on the information given in the problem.

(b) Find the derivative of that function to get the critical number.

(c) Test the C.N. using the first or second derivative test.

(d) Answer any question given in the problem.

Example 4: A manufacturer sells 500 units per week at 31 dollars per unit. If the price is reduced by one dollar, 20 more units will be sold.

To maximize the revenue, find:

(a) the selling price let x be the number of one dollar reduction price in dollars per unit: 31 - x

(b) the number of units sold number of units sold: 500 + 20x

(c) the maximum revenue Revenue = (price per unit) . (number of units) R(x) = (31 - x) . (500 + 20x) = -20x2 + 120x + 15500

Once the equation of the revenue function is found, use the first derivative to find C.N., test the C.N. using the first or second derivative and then answer the following questions: (1) find the selling price to maximize the revenue; (2) find the number of units sold to maximize the revenue; (3) find the maximum revenue.

R (x) = -40x + 120 = R (x) = 0 = -40x + 120 = 0 = x = 3

Test the critical number x = 3 with the second derivative: R (x) = -40 < 0 , relative maximum

(1) the selling price to maximize the revenue is 31 - 3 = 28 dollars per unit (2) the number of units sold to maximize the revenue is 500 + 20(3) = 560 units (3) the maximum revenue is R(3) = (28) . (560) = $15 680

Math 201-103-RE - Calculus I

Application of the derivative (2) Business and Economics

Page 4 of 15

Example 5: A rectangular tennis court of 1800 square meters is to be fenced with 2 types of materials. The shorter sides are made with fence material costing $100 per meter and the other sides with fence material costing $50 per meter. Find the dimensions of the court to minimize cost.

Let x be the shorter side and y be the other side in meter.

Total cost of the fence: C = 2x(100) + 2y(50) = 200x + 100y

Area:

1800 = x . y

=

y=

1800 x

replace y in the equation of C :

1800

180 000

C(x) = 200x + 100 = 200x +

x

x

C

(x)

=

200

-

180 000 x2

=

C (x) = 0

=

200

-

180 000 x2

=

0

=

x = 30

Test the critical number

x = 30

with the second derivative:

C

360 000 (x) = x3

=

C

(30) > 0 , relative minimum

To minimize the cost, the dimensions of the court should be

x = 30 meters by

y=

1800 30

= 60 meters.

Example 6: A company has established that the revenue function in dollars is R(x) = 2x3 + 40x2 + 8x and the cost function in dollars is C(x) = 3x3 + 19x2 + 80x - 800 .

Find the price per unit to maximize the profit.

To maximize the profit, need the profit function P = R - C

P (x) = (2x3 + 40x2 + 8x) - (3x3 + 19x2 + 80x - 800) = -x3 + 21x2 - 72x + 800

Once the equation of the profit function is found, use the first derivative to find C.N., test the C.N. using the first or second derivative and then answer the question:

P (x) = -3x2 + 42x - 72 = P (x) = 0 = -3x2 + 42x - 72 = 0 = -3(x - 2)(x - 12) = 0

the critical numbers are: x = 2 ; x = 12

use second derivative test: f (x) = -6x + 42

x = 2 = f (2) > 0 = relative minimum ; x = 12 = f (12) < 0 = relative maximum

To maximize the profit, use x = 12 . To find the price per unit:

p = R = 2x3 + 40x2 + 8x = 2x2 + 40x + 8

x

x

at x = 12 = p(12) = 2(12)2 + 40(12) + 8 = $776 per unit.

Math 201-103-RE - Calculus I

Application of the derivative (2) Business and Economics

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Price Elasticity of Demand % change in quantity

Definition: price elasticity of demand is a ratio of: % change in price

Notation and formula of price elasticity of demand:

Symbol of price elasticity of demand is the Greek letter pronounced "ETA"

Formula of price elasticity of demand:

=

x

p .p

The proof is given in class.

where p is the demand function (price per unit) ; x is the demanded quantity (number of units) and p is the derivative of p .

Price Elasticity of Demand Interpretation:

if || > 1 < -1 or > 1 - demand is elastic

if || < 1 -1 < < 1 - demand is inelastic

if || = 1 = ?1 - demand has unit elasticity

Elastic demand - decrease price implies increase revenue. Inelastic demand - decrease price implies decrease revenue. Unit elasticity of demand - decrease price implies unchanged revenue.

Example 7: The demand of a product is p = 25 - x2 where x is the demanded quantity. Find: (a) the price elasticity of demand.

Use formula = p ; need to replace p = 25 - x2 and p = -2x x.p

25 - x2 x2 - 25 = x . (-2x) = 2x2

(b) the intervals of elasticity

Two equations to solve for x :

x2 - 25 2x2 = 1

=

x2 - 25 = 2x2

=

-25 = x2

=

no solution

x2 - 25 2x2

=

-1

=

x2 - 25 = -2x2

=

3x2 - 25 = 0

=

x2

=

25 3

=

x ?2.89

Next page, testing on the number line with x = 2.89 and x = 0 to x = 5 since p cannot be negative and cannot exceed x = 5 .

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