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Math 201-103-RE - Calculus I Business Functions

Application of the derivative (2) Business and Economics

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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 ina 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 .

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) = The derivative of the cost function C(x) is called marginal cost with notation: C ′ (x)= dC dx dP dx dR dx

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

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 dR dx = −9(300)2 + 1200(300) =−450 000
x=300

marginal revenue at x = 300 =⇒ R′ (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 dR dx = −9(100)2 + 1200(100) = 30 000
x=100marginal revenue at x = 100 =⇒ R′ (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

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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 marginal cost: C ′ (x) = dC = 714x + 1800 dx dC marginal cost at x = 50 =⇒ C ′ (50) = dx

= 714(50) + 1800 = 37 500
x=50

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

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 = −9x2 + 486x − 1800 dx dP dx = −9(10)2 + 486(10) − 1800 = 3300
x=10

marginal profit at x = 10 =⇒ P ′ (10) =

Interpretation: If production increases from 10 to 11 units, the profit increases by 3300dollars. (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 dP dx = −9(100)2 + 486(100) − 1800 = −750 000
x=100

marginal profit at x = 100 =⇒ P ′ (100) =

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

Math 201-103-RE - Calculus I

Application of the derivative (2) Business and Economics

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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 theinformation 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...