### Introduction to Calculus in Economics (continued):In the previous Problem Set question, we started looking at the cost function $C(x)$, the cost of a firm producing $x$ items. An important microeconomics concept is the **marginal cost**, defined in (non-mathematical introductory) economics as the cost of producing one additional item.If the current production level is $x$ items with cost $C(x)$, then the cost of computing $h$ additional items is $C(x + h)$. The average cost of those $h$ items is $\frac{C(x+h) - C(x)}{h}$. As we analyze the cost of just the last item produced, this can be made into a mathematical model by taking the limit as $h \rightarrow 0$, i.e. the derivative $C'(x)$. Use this function in the model below for the Marginal Cost function $MC(x)$.### Problem Set question:The cost, in dollars, of producing $x$ units of a certain item is given by\[ C(x) = 0.04x^3 - 5x + 500. \](a) Find the marginal cost function.\[ MC(x) = \](b) Find the marginal cost when 40 units of the item are produced.The marginal cost when 40 units are produced is $ \[ \text{Number} \].(c) Find the actual cost of increasing production from 40 units to 41 units.The actual cost of increasing production from 40 units to 41 units is $ \[ \text{Number} \].

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Answered: Introduction to Calculus in Economics…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter6: Applications Of The Derivative

Section6.EA: Extended Application A Total Cost Model For A Traning Program

Problem 1EA: In this application, we set up a mathematical model for determining the total costs in setting up a...

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In this application, we set up a mathematical model for determining the total costs in setting up a training program, such as a hospital might use. Then we use calculus to find the time interval between training programs that produces the minimum total cost. The model assumes that the demand for trainees is constant and that the fixed cost of training a batch of trainees is known. Also, it is assumed that people who are trained, but for whom no job is readily available, will be paid a fixed amount per month while waiting for a job to open up. The model uses the following variables. D = demand for trainees per month N = number of trainees per batch C1 = Fixed cost of training a batch of trainees C2 = marginal cost of training per trainee per month C3 = salary paid monthly to a trainee who has not yet been given a job after training m = time interval in months between successive batches of trainees t = length of training program in months Z(m) = total monthly cost of program The total cost of training a batch of trainees is given by C1+NtC2. However, N=mD, so that the total cost per batch is C1+mDtC2. After training, personnel are given jobs at the rate of D per month. Thus, ND of the trainees will not get a job the first month, N2D will not get a job the second month, and so on. The ND trainees who do not get a job the first month produce total costs of (ND)C3, and so on. Since N=mD, the costs during the first month can be written as. (ND)C3=(mDD)C3=(m1)DC3 While the costs during the second month are (m2)DC3, and so on. The total cost for keeping the trainees without a job is thus (m1)DC3+(m2)DC3+(m3)DC3++2DC3+DC3 Which can be factored to give DC3[(m1)+(m2)+(m3)++2+1] The expression in bracket is the sum of the terms of an arithmetic sequence, discussed in most algebra texts. Using formulas for arithmetic sequences, the expression in brackets can be shown to equal m(m1)2, so that we have DC3[m(m1)2](1) As the total cost for keeping jobless trainees. The total cost per batch is the sum of the training cost per batch. C1+mDtC2, and the cost of keeping trainees without a proper job, given by equation 1, since we assume that a batch of trainees is trained every m months, the total cost per month, Z(m), is given by Z(m)=C1+mDtC2m+DC3[m(m1)2]m=C1m+DtC2+DC3(m12) Source: P.I., Goyal and S.K. Goyal. Find Z(m)

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