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# Integration: Cost functions from marginal cost functions

If C is the cost of producing an output x, then marginal cost function MC = dc/dx

Cost functions from marginal cost functions

If C is the cost of producing an output x, then marginal cost function MC = dc/dx

Using integration, as the reverse process of differentiation, we obtain,

Cost function C = Ōł½ ( MC ) dx + k

Where k is the constant of integration which is to be evaluated,

Average cost function AC = C/X, x ŌēĀ 0

Example 3.9

The marginal cost function of manufacturing x shoes is 6 + 10x ŌłÆ 6x2 . The cost producing a pair of shoes is Ōé╣12. Find the total and average cost function.

Solution:

Given,

Marginal cost MC = 6 + 10x ŌłÆ 6x2

C = Ōł½ MC dx + k

= Ōł½( 6 + 10x ŌłÆ 6x2 ) dx + k

= 6x + 5x2 ŌłÆ 2x 3 + k (1)

When x= 2 , C = 12 (given)

12 = 12 + 20 ŌłÆ16 + k

k=-4

C = 6x + 5x2 ŌłÆ 2x3 ŌĆō 4 Example 3.10

A company has determined that the marginal cost function for a product of a particular commodity is given by MC = 125 + 10x ŌłÆ x2/9 where C rupees is the cost of 9 producing x units of the commodity. If the fixed cost is Ōé╣250 what is the cost of producing 15 units.

Solution: Example 3.11

The marginal cost function MC = 2 + 5ex (i) Find C if C (0)=100 (ii) Find AC.

Solution: Rate of growth or sale

If the rate of growth or sale of a function is a known function of t say f(t) where t is a time measure, then total growth (or) sale of a product over a time period t is given by,

Total sale = Example 3.12

The rate of new product is given by f (x) = 100 ŌłÆ 90 eŌłÆx where x is the number of days the product is on the market. Find the total sale during the first four days. (eŌĆō4=0.018)

Solution: Example 3.13

A company produces 50,000 units per week with 200 workers. The rate of change of productions with respect to the change in the number of additional labour x is represented as 300 ŌłÆ 5x 2/3 . If 64 additional labours are employed, find out the additional number of units, the company can produce.

Solution:

Let p be the additional product produced for additional of x labour, Ōł┤ The number of additional units produced 16128

Total number of units produced by 264 workers

50,000 + 16,128 = 66128 units

Example 3.14

The rate of change of sales of a company after an advertisement campaign is represented as, f (t ) = 3000e ŌłÆ0.3t where t represents the number of months after the advertisement. Find out the total cumulative sales after 4 months and the sales during the fifth month. Also find out the total sales due to the advertisement campaign e ŌłÆ1 . 2 = 0.3012, eŌłÆ1.5 = 0.2231 .

Solution: Example 3.15

The price of a machine is 6,40,000 if the rate of cost saving is represented by the function f(t) = 20,000 t. Find out the number of years required to recoup the cost of the function.

Solution:

Saving Cost          S(t) = Ōł½t0 20000t dt

= 10000 t2

To recoup the total price,

10000 t2 = 640000

t2 = 64

t = 8

When t = 8 years, one can recoup the price.

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12th Business Maths and Statistics : Chapter 3 : Integral Calculus - II : Integration: Cost functions from marginal cost functions | Example Solved Problems with Answer, Solution, Formula