Answered: If the Marginal Cost for a product is…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Question

If the Marginal Cost for a product is found to be MC= 1500/15x+100' , and the cost of 80 units is found to be $6000, what are:

the Total Costs?

the Fixed Costs?

Expert Solution

Step 1

We have,

x= 80

Total cost = TC = $6000 at x = 80

Marginal cost,

MC = 1500/ (15x+100)

To find the total cost function, integrate MC function.

$\begin{array}{rcl} T C & = & \int M C (x) d x \\ = & \int \frac{1500}{15 x + 100} d x \end{array}$

Let, t = 15x+100

$\begin{array}{rcl} \frac{d t}{d x} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 15 \end{array} \begin{array}{rcl} \end{array} d x = \begin{array}{rcl} \frac{1}{15} \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} t \end{array} S \begin{array}{rcl} o \end{array} \begin{array}{rcl} , \end{array} T C = \begin{array}{rcl} \int \end{array} \begin{array}{rcl} \frac{1500}{t} \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} \frac{1}{15} \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} t \end{array} T C \begin{array}{rcl} = \end{array} \begin{array}{rcl} \int \end{array} \begin{array}{rcl} \frac{100}{t} \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} t \end{array} T C \begin{array}{rcl} = \end{array} \begin{array}{rcl} 100 \end{array} \begin{array}{rcl} \int \end{array} \begin{array}{rcl} \frac{1}{t} \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} t \end{array} T C \begin{array}{rcl} = \end{array} \begin{array}{rcl} 100 \end{array} \begin{array}{rcl} [\ln (t)] \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} c \end{array} T C \begin{array}{rcl} = \end{array} \begin{array}{rcl} 100 \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} l \end{array} \begin{array}{rcl} n \end{array} \begin{array}{rcl} (15 x + 100) \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} c \end{array}$

Step 2

If x = 80,

TC = 6000

$100 \ln (15 x + 100) + c = 6000 100 \ln (15 (80) + 100) + c = 6000 100 \ln (1300) + c = 6000 311.4 + c = 6000 c = 5688.6$

Put the value of c in TC function to get complete TC function.

TC (x) = 100 ln (15x+100) +5688.6

At x= 80, TC = $6000

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