• Author / Uploaded
• Ted Bryan D. Cabagua

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APPLICATIONS OF DERIVATIVES IN BUSINESS 1. Let y = C(x), where y represents the cost to manufacture x items. Thus, C(x) is described as the cost function. Example: The total cost in thousands of Pesos to manufacture x electric generators is given by C(x) = -x3 + 15x2 +1000 2. Average cost per item, AC(x), is determined by dividing the total cost by the number of items. AC(x) = C(x)/ x In the given example where C(x) = -x3 + 15x2 +1000, AC(x) = -x2 + 15x + 1000/x 3. Marginal Cost (MC) gives the rate of change in the cost of producing one more item after x items have already been produced. It is the approximate change in the cost resulting from one additional unit of output. It is the first derivative of the of the cost function. MC = C’(x) Example 1: The cost to produce x units of an item is given by C(x) = 9x 2 – 4x + 8. a. Find the average cost per unit to produce: 10 units, 20 units AC(x) = C(x)/x = 9x – 4 + 8/x C(10) = 9(10) – 4 + 8/10 = 86.8 C(20) = 9(20) – 4 + 8/20 = 176.4 b. Find the 1st derivative of the average cost function AC’(x) = 9 – 8/x2 = (9x2 – 8)/x2 Example 2: Given the cost function C(x) = 2x 2 + 5x + 18, find the minimum average cost. a. Solve for the average cost, AC(x). AC(x) = 2x + 5 + 18/x b. Solve for AC’(x). AC’(x) = 2 – 18/x2 c. Equate AC’(x) to ‘0’ and solve for x. 2 – 18/x2 =0

(2x2 – 18)/x 2 =0 2(x2 – 9) =0 x=3 x = -3 d. Only x = 3 since it is positive and since C”(x) = 4 is positive, x = 3 is a minima. e. Substitute x = 3 to AC(x) = 2x + 5 + 18/x AC(3) = 2(3) + 5 + 18/3 = 17 Therefore, the minimum average cost is P17 Example 3: Suppose that the total cost (in hundreds of Pesos) to produce x (in thousand units) of an item is given by the function C(x) = 4x2 + 100x + 500. Find the marginal cost for the following values of x: a. x = 5 or (5000 units) Cost for x = 5 or (5000 units) is C(x) = 4x2 + 100x + 500 C(5) = 4(5)2 + (100 x 5) + 500 = P1100 or P110,000 (P1100 x 100) Average cost per item for x = 5 or (5000 units) is AC(x) = 4x + 100 + 500/x AC(5) = (4x5) + 100 + (500/5) = P220 or P22,000 (P220 x 100) Marginal cost MC = C’(x) = 8x + 100 C’(5) = (8 x 5) + 100 = P140 or P14,000 (P140 x 100) Therefore, after 5000 units of an item have been produced, the cost to produce 1000 more units will be approximately 140 (in hundreds of pesos) or P14,000. b. x = 30 or (30000 units) Cost for x = 30 or (30000 units) is C(x) = 4x2 + 100x + 500 C(30)= 4(30) 2 + (100x30) + 500 = P7100 or P710,000 (P7100 x 100) Average cost per item for x = 30 or (30000 units) is AC(x) = 4x + 100 + 500/x AC(30) = (4x30) + 100 + (500/30)

= P131.67 or P13,167 (P131.67 x 100) Marginal cost MC = C’(x) = 8x + 100 C’(30) = (8 x 30) + 100 = P340 or P34,000 (P340 x 100) Therefore, after 30000 units of an item have been produced, the cost to produce 1000 more units will be approximately P340 (in hundreds of pesos) or P34,000. 4. When AC’(x) = 0. The critical values (value of x) may either be considered as the minimum number of units required to achieve the minimum cost or the maximum number of units required to achieve the maximum cost. If C”(x) is negative, x is maximum value. If C”(x) is positive, x is a minimum value. Example 4: In the previous example, when AC’(x) = 0, x is computed as AC’(x) = 4 – 500/x2 4 – 500/x2 = 0 (4x2 – 500)/x2 = 0 4x2 – 500 = 0; therefore 0 = 4(x2 – 125) where x =

√ 125

or 11

X is approximately 11 or (11000 units) Since C”(x) = 8 is positive, x = 11 is the minimum value. The minimum cost is determined by substituting the value of x to AC(x). AC(x) = 4x + 100 + 500/x AC(11) = (4 x 11) + 100 + (500/11) = P189.45 or P18,945 (P189.45 x 10) 5. Revenue (R) – For any given demand function y = f(x), the total Revenue function, R, is the product of x (the number of units demanded) and y (the price per unit demanded). R=x●y

or

R = x ● f(x)

6. Marginal Revenue function (MR) – rate of change (derivative) of the total Peso vale received with the total number of units sold. It is also the approximate change in revenue that results from selling one additional unit. MR = R’=

x●

d f (x) dx

+ f(x)

Example: The revenue for selling x units of an item is given by R(x) = 800x + 10x2. Find the Marginal Revenue when a. x = 10, b = 20, c = 32 R(x) = 800x + 10x2 R’(x) = 800 + 20x a. R’(10) = 800 + 20(10) = 1000 b. R’(20) = 800 + 20(20) = 1200 c. R’(32) = 800 + 20(32) = 1440 Example: The demand function of a commodity is given by the function y = 200 – 5x where x is the quantity demand and y is the unit price. Determine the price and quantity for which revenue is maximum. Demand: Revenue

y = 200 – 5x R=x●y = x (200 – 5x) = 200x – 5x2 R’ = 200 – 10x

To get the critical value/s, equate R’ to 0 and solve for x 200 – 10 x = 0 x = 20 Substitute x to the demand function, y = 200 – 5(20) = 100 To determine if x = 20 is a minima or maxima, get the 2nd derivative of R. If R’ is positive, x = 20 is a minima. If R’ is negative, x = 20 is a maxima. R” = -10 and since it is negative, x = 20 is a maxima. Therefore a maximum revenue is expected for a sale of 20 units at P100 each. Example: The price per bar of a chocolate candy is given by the function y=

6 x − 5 250

How many bars must be sold to maximize revenue? What is the maximum revenue? Revenue:

R=x●y =x

6 x ( − ) 5 250

= 6x/5 – x2/250 R’ = 6/5 – 2x/250 = (300 – 2x)/250

To get the critical value/s, equate R’ to 0 and solve for x (300 – 2x)/250 = 0 300 – 2x = 0 x= 150 Substitute x = 150 to the revenue function, R = 6(150)/5 – 1502/250 = 90 To determine if x = 150 is a minima or maxima, get the 2 nd derivative of R. If R’ is positive, x = 20 is a minima. If R’ is negative, x = 20 is a maxima. R” = 2/150 and since it is positive, x = 150 is a minima. Therefore a maximum revenue of P90 is expected for a sale of 150 bars.

7. The Profit Function is the difference of the revenue function, R(x) and the cost function, C(x). P(x) = R(x) – C(x) Maximize profit may be achieved when the 1st derivative of the Profit function is equated to 0. Example:

A certain item can be produced at a unit cost of P10. The demand function for the product is D = 90 - .02x, where D is the price in pesos, and x is the number of units. a. How many units must be produced to maximize Profit? b. What is the price that gives maximum profit? c. What is the maximum profit? Cost Function C(x) = 10x Revenue function R(x) = x(D) = x(90 – .02x) = 90x – .02x2 Profit function P(x) = R(x) – C(x) = (90x – .02x2) – 10x = 80x – .02x2 Maximum profit may be achieved when 1 st derivative of P(x) = 0 P(x) = 80x – .02x2 P’(x) = 80 – .04x At P’(x) = 0, 80 – .04x = 0

x = 2000 Test for P”(x) to check if x = 2000 is maximum or minimum. P’(x) = 80 – .04x P”(x) = – .04. Since P”(x) is negative x = 2000 is maximum. The price that gives maximum profit is determined by substituting x to the Demand function D = 90 - .02x. D = 90 - .02(2000) = P50 The maximum profit is determined by substituting x to the Profit function P(x) = 80x – .02x2 P(2000) = 80(2000) - .02(2000)2 = 160,000 - . 02(4,000,000) = 160,000 – 80,000 = P80,000