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Sample Internal Assessment
Sample Internal Assessment

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Table of content
Rationale
Aim
Background
Methodology
Data collection and calculation
Conclusion
Reflection
Bibliography

To investigate the growth of Eastman Auto and Power Limited Company in terms of profit with respect to time and cost of goods sold over a period of 5 years and determination of the corresponding time and expenditure to maximize the profit.

To investigate the growth of Eastman Auto and Power Limited Company in terms of profit with respect to time and cost of goods sold over a period of 5 years and determination of the corresponding time and expenditure to maximize the profit. Reading Time
10 mins Read
To investigate the growth of Eastman Auto and Power Limited Company in terms of profit with respect to time and cost of goods sold over a period of 5 years and determination of the corresponding time and expenditure to maximize the profit. Word Count
1,916 Words
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Table of content

Rationale

Mathematics is a very versatile field. It has always inspired me go beyond the scope of the curriculum and analyze different real life scenarios using mathematics. It is extensively used in the study of businesses. Being and inquirer and aspiring to be an entrepreneur in the future, I wanted to effectively study businesses with the help of mathematics. The concepts of maxima, minima, integration and differentiation have helped me analyze the data to arrive at useful conclusions.

 

Before the pandemic hit the businesses, I wanted to study the growth of a company to determine the time and expenditure to maximize the profit. As I aspire to be an entrepreneur, it’s essential to understand the growth of a company and on which factors and how it depends. I have taken the data from the yearly balance sheets of the company Eastman Auto and Power limited over a period of approximately 5 years from June, 2016 to March, 2020. To determine how the profit varies with respect to time and cost of goods and to determine the maximum profit, the concepts of maxima, minima, integration and differentiation appeared to be very useful.

 

The study would be useful to the businesses to take calculated risks and right decisions to grow it. It would be also helpful to maximize the profit and determine the conditions under which it attains so.

Aim

To investigate the growth of Eastman Auto and Power limited company in terms of profit with respect to time and cost of goods sold over a period of 5 years and thus determining the corresponding time and expenditure to maximize the profit.

Background

Profit and cost of goods sold

The immediate capital earned from selling goods produced by a company is the total revenue. The company uses a portion of the capital earned or the revenue to produce further goods or cuts the costs for making the goods (which is the Cost of goods sold) from the revenue earned. The capital which remains is called the profit.

Differentiation

Let y = f(x) be a single valued function of x defined in some interval. Let x be any value of x in the domain of definition of the function and the corresponding value of y is y = f(x). Suppose for an increment ∆x of x the corresponding increment in y is ∆y.

 

Then,\(\frac{∆y}{∆x}=\frac{dy}{dx}=\frac{f(x+h)-f(x)}{h}\) is called the derivative of the function y with respect to x, provided the limit exists.

Integration

Let f(x) be a bounded single valued continuous function defined in the interval [a,b]. The interval [a,b] is divided into n equal sub-intervals, each of length h.

 

Then,\(\displaystyle\int\limits^b_af(x)dx=h\sum^{n-1}_{r=0}f(a+rh)\) is the integration or the integral of the function f(x) with respect to x between the limits a and b.

Concepts of minima and maxima

The maximum value of any function y = f (x) is called the maxima of the function, and the minimum value of the function is called the minima of the function. This is an application of differential calculus. Any function y = f (x) could be graphically represented in Cartesian coordinate system. The slope at any point of the function could be determined by finding the derivative of the function at the particular point. Slope of any curve equal to zero signifies that the tangent at that point of the curve is parallel to X – axis. As a result, if the equation of slope of the curve is equated to zero, and the root of the equation is determined, then it would be the value of x at which the function is either maximum or minimum. In order to determine the maxima or minima, the equation of slope is further differentiated and if the value of equation is obtained to be negative, then the point is maxima, and vice versa.

Methodology

The study has been done mainly in three steps. Firstly, the values of Profit (taken quarterly) with respect to time has been plotted to obtain the Profit as a function of time and the local maximum has been determined to know at what time the Profit reaches its maximum value. Secondly, the values Cost of goods sold (COGS) (taken quarterly) with respect to time has been plotted to obtain the Cost of goods sold (COGS) as a function of time and the local minimum has been determined to know at what time the Cost of goods sold (COGS) reaches its minimum value. Lastly, the Profit has been expressed as a function of Cost of goods sold (COGS) to determine how the Profit varies with respect to the Cost of goods sold (COGS) and to determine the local maximum or the maximum value of the Profit and at what value it reaches so.

Data collection and calculation

Year
Month
Index
Profit
2016
June
1
58780477
2016
September
2
61552683
2016
December
3
64854311
2017
March
4
68654782
2017
June
5
72439854
2017
September
6
68610768
2017
December
7
65005376
2018
March
8
54414169
2018
June
9
50446665
2018
September
10
56509717
2018
December
11
60286127
2019
March
12
50611648
2019
June
13
45501624
2019
September
14
42855761
2019
December
15
43477561
2020
March
16
32288409
Figure 1 - Table On Variation Of Profit (Taken Quarterly) From June, 2016 To March, 2020
Figure 2 - Variation Of Yearly Profit (Taken Quarterly) From The June, 2016 To March, 2020
Figure 2 - Variation Of Yearly Profit (Taken Quarterly) From The June, 2016 To March, 2020

After plotting the values of profit (taken quarterly for each year) with respect to time (as per indexing), we get the profit expressed as a function of time. The equation is given by, y = - 22820x+ 2000000x + 60000000…(i) where y = p (say) denotes the profit and x = t (say) denotes the time.

 

∴p = - 22820t+ 2000000t + 60000000…(i)

Calculation of local maximum

Now, to calculate at what time the profit reaches its maximum value, we will calculate the local maximum of the function and at what point it reaches so. To calculate the maximum of the function p = f (t), we have to first equate the derivative of the function to zero. 

 

p = - 22820t+ 2000000t + 60000000…(i)

 

\(\frac{dp}{dt}=\frac{d}{dt}\) (-22820t+ 2000000t + 60000000)

 

\(\frac{dp}{dt}=\frac{d}{dt}\big(-22820t^2\big)+\frac{d}{dt}(2000000t)\frac{d}{dt}(60000000)\)

 

\(\frac{dp}{dt}\)= - 45640t + 2000000 + 0

 

\(\frac{dp}{dt}\) = - 45640t + 2000000…(ii)

 

Now, we equate the equation obtained to zero.

 

\(\frac{dp}{dt}\) = - 45640t + 2000000 = 0

 

\(t=\frac{-2000000}{-45640}\)

 

t = 43.82

 

Now, we will further differentiate the equation and calculate its numerical value at t = 43.82

 

\(\frac{dp}{dt}\) = - 45640t + 2000000

 

\(\frac{d^2p}{dt^2}\) = - 45640

 

Therefore, at the point t = 43.82,

 

\(\frac{d^2p}{dt^2}\) = - 45640, which is negative

 

Hence, p = - 22820t2 + 2000000t + 60000000 will have its maximum value at t = 43.82. At t = 43.82, the value of the local maximum is,

 

p = - 22820t2 + 2000000t + 60000000

 

p = - 22820×43.82+ 2000000×43.82 + 60000000

 

p = 103821209.43 (in INR)

Figure 3 - Variation Of Cost Of Goods Sold (COGS) (Taken Quarterly) From June, 2016 To March, 2020
Figure 3 - Variation Of Cost Of Goods Sold (COGS) (Taken Quarterly) From June, 2016 To March, 2020
Year
Month
Index
COGS
2016
June
1
1349071587
2016
September
2
767738939
2016
December
3
1356275870
2017
March
4
1590624659
2017
June
5
1620682736
2017
September
6
1535555852
2017
December
7
1615635389
2018
March
8
1946676706
2018
June
9
1667878045
2018
September
10
2137319305
2018
December
11
2114832764
2019
March
12
2461816607
2019
June
13
2707886937
2019
September
14
2660489057
2019
December
15
3137073824
2020
March
16
3704206763
Figure 4 - Table On Variation Of Cost Of Goods Sold (COGS) (Taken Quarterly) From June, 2016 To March, 2020

After plotting the values of the Cost of goods sold (COGS) (taken quarterly for each year) with respect to time (as per indexing), we get the profit expressed as a function of time. The equation is given by, y = 80000000x+ 600000000x + 1000000000…(iii) where y = c (say) denotes the profit and x = t (say) denotes the time.

 

∴c = 80000000t2 + 600000000t + 1000000000…(iii)

Calculation of local minimum

Now, to calculate at what time the Cost of goods sold (COGS) reaches its minimum value, we will calculate the local minimum of the function and at what point it reaches so. To calculate the minimum of the function c = f (t), we have to first equate the derivative of the function to zero.

 

c = 80000000t2 + 600000000t + 1000000000…(iii)

 

\(\frac{dc}{dt}=\frac{d}{dt}\) (80000000t2 + 600000000t + 1000000000)

 

\(\frac{dc}{dt}=\frac{d}{dt}\big(80000000t^2\big)+\frac{d}{dt}(600000000t)+\frac{d}{dt}(1000000000)\) 

 

\(\frac{dc}{dt}\) =160000000t + 600000000…(iv)

 

Now, we equate the equation obtained to zero.

 

\(\frac{dc}{dt}\)=160000000t + 600000000 = 0

 

\(t=\frac{-600000000}{160000000}\)

 

t = 3.75

 

Now, we will further differentiate the equation and calculate its numerical value at t = 3.75

 

\(\frac{dc}{dt}\) = 160000000t + 10000000

 

\(\frac{d^2p}{dt^2}\) = 160000000

 

Therefore, at the point t = 3.75,

 

\(\frac{d^2p}{dt^2}\) = 160000000, which is positive.

 

Hence, c = 80000000t2 + 600000000t + 1000000000 will have its minimum value at t = 3.75.

 

At t = 3.75, the value of the local minimum is,

 

c = 80000000t2 + 600000000t + 1000000000

 

c = 80000000×3.752 + 600000000×3.75 + 1000000000

 

c = 1337500000(in INR)

 

Now, dividing equation (ii) by equation (iv) we get,

 

\(\frac{dp}{dt}\) = - 45640t + 2000000…(ii)

 

\(\frac{dc}{dt}\) = 160000000t + 600000000…(iv)

 

\(\frac{\frac{dp}{dt}}{\frac{dc}{dt}}=\frac{-45640t+2000000}{160000000t+600000000}\)

 

\(\frac{dp}{dc}=\frac{-0.4564t+20}{1600t+6000}\)

 

\(\frac{dp}{dc}\) = - 0.00028 - \(\frac{21.71}{1600t+6000}\) …(v)

 

We get equation (v) which is the general equation for the rate of change of Profit with respect to Cost of goods sold.

 

Now, we put the value of t in terms of c, which we can calculate from equation (v).

 

c = 8000000t2 + 600000000t + 1000000000…(iii)

 

\(\frac{c}{10^8}\) = 0.08t2 + 6t + 10

 

\(t =\frac{-c}{71.59\times10^8}\) -73.29

 

Now, by putting the value of t in equation (v) we get,

 

\(\frac{dp}{dc}=-0.00028-\frac{{21.71}{-c}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}\)…(vi)

 

Now, integrating equation (vi) we get,

 

\(\frac{dp}{dc}=-0.00028-\frac{{21.71}{-c}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}\)

 

\(\displaystyle\int dp=\displaystyle\int(-0.00028-\frac{{21.71}{-c}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000})dc\)

 

\(p=\displaystyle\int-0.00025dc-\displaystyle\int\frac{{21.71}{-c}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}dc\)

 

\(p=\displaystyle\int-0.000028dc-\displaystyle\int\frac{{21.71}{-c}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}dc\)

 

p = - \(\frac{7c}{25000}+\frac{388554725ln(c+497836860000)}{4}\) + C1

 

Where C1 is the constant of integration.

 

Now, to calculate the numerical value of the constant of integration, we put the values of Profit and Cost of goods sold when the time (as per indexing) is 1.

 

p = - \(\frac{7c}{25000}+\frac{388554725ln(c+497836860000)}{4}\) + C1

 

Putting p = 58780477 and c = 1349071587 we get,

 

58780477 = - \(\frac{7×1349071587}{25000}+\frac{388554725ln(1349071587+497836860000)}{4}\) + C1

 

C1 = 58780477 + 377740.04 + 97138681.25×26.936

 

C1 = 2675685735.19

 

Therefore, we get the Profit expressed as a function of Cost of goods sold as,

 

p = - \(\frac{7c}{25000}+\frac{388554725ln(c+497836860000)}{4}\) + 2675685735.19…(vii)

Figure 5 - Variation Of Profit With Respect To Cost Of Goods Sold
Figure 5 - Variation Of Profit With Respect To Cost Of Goods Sold

Calculation of local maximum

Now, to find out the local maximum or local minimum of the above function p = f(c), we first equate the derivative of the function to zero.

 

\(\frac{dp}{dc}=-0.00028-\frac{{21.71}{}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}\) = 0

 

\(\frac{{-21.71}{}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}\) = 0.00028

 

\(\frac{-21.71}{0.000280}-6000=1600\bigg(\frac{-c}{71.59×10^8}-73.29\bigg)\)

 

c = 844760721607

 

Now, we will further differentiate the equation and calculate its numerical value at c = 844760721607

 

\(\frac{dp}{dc}=-0.00028-\frac{{21.71}{}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}\)

 

\(\frac{d^2p}{dc^2}=\frac{d}{dc}(-0.00028-\frac{{21.71}{}}{1600(\frac{-c}{71.59\times10^8}-73.29)+6000}\big)\)

 

\(\frac{d^2p}{dc^2}=\frac{d}{dc}(-0.00028)-\frac{d}{dc}(\frac{{21.71}{}}{1600\bigg(\frac{-c}{71.59\times10^8}-73.29\bigg)+6000})\)

 

\(\frac{d^2p}{dc^2}=\frac{-388554725}{4 (c+497836850000)^2}\)

 

Therefore, at the point c = 844760721607,

 

\(\frac{d^2p}{dc^2}\) = - 5.39×10-17, which is negative

 

Hence, p = - \(\frac{7c}{25000}+\frac{388554725ln(c+497836860000)}{4}\) + 2675685735.19 will have its local maximum at c = 844760721607.

 

At c=844760721607, the value of the local maximum is,

 

p = - \(\frac{7c}{25000}+\frac{388554725ln(c+497836860000)}{4}\) + 2675685735.19

 

p = - \(\frac{7×844760721607}{25000}+\frac{388554725ln(844760721607+497836860000)}{4}\) + 2675685735.19

 

p = 5152000000 (in INR)

 

Therefore, the Profit is maximum which is 5152000000 (in INR) when the Cost of goods sold is 84460721607 (in INR).

Conclusion

According to the data taken regarding the Profit and Cost of goods sold (taken quarterly for each year) from the year 2016 to the beginning of the year 2020, the following assumptions can be made:

  • From the Profit vs. Time (as per indexing) graph, the values of Profit keeps going with increasing time and reaches its maximum at time t = 43.82 but the values of Cost of goods sold have gone up with each passing quarter. This indicates inefficiency of the production work or the deterioration of the quality of goods or the sales has gone significantly down.
  • From the Cost of goods sold (COGS) vs. Time (as per indexing) graph, the values of Cost of goods sold have gone up with each passing quarter. The value of the Cost of goods sold (COGS) reaches its minimum value at time t = 3.75. The values of Profit have not increased as such. This indicates that the production cost has increased significantly over time due inflation or significant increase in the quantity of goods produced.
  • From the Profit vs. Cost of goods (COGS) graph, the profit almost remains constant and reaches its maximum 5152000000 (in INR) when Cost of goods sold (COGS) c is 84460721607 (in INR). This indicates that the sales have significantly gone down or the quality of the products has deteriorated.
  • The company should invest more on checking the quality of the goods than producing more goods.
  • The company should invest more on the marketing department in order to reach the people and make them aware of its various goods.

Reflection

Strength

  • The data taken, contains the values of Profit, Cost of goods sold (COGS), which are taken quarterly for each year, varies over a period of approximately 4 years, starting from June, 2016 to March, 2020. The variations of Profit and Cost of goods sold with respect to time have been studied. The variation of Profit with respect to the Cost of goods sold has also been studied, which makes the study coherent.
  • The values of Profit and Cost of goods sold has been taken quarterly for each year instead of just considering the net yearly profit, which makes the study more coherent and relevant to arrive at certain important conclusions.
  • The data has been taken from the balance sheets of the company which makes it authentic, error free and could be considered to do a critical analysis.

Weakness

  • The data has been taken from the balance sheets of a registered company but there is little to no scope of verifying the data, which may lead to discrepancy in the collected data which would make the study less coherent.
  • It was assumed that the variation of Profit and Cost of goods sold would follow the same trend in the future to arrive at certain conclusions. However, this might not be the case in real life.
  • Other real life factors related to the working of the company which actually effects the profit and loss statement and the cost of goods have not been considered which makes the study less coherent and cannot be relied upon entirely to take important decisions for the growth the company.

Further scope

Mathematics is essential to analyze various aspects and the balance sheets of a company to make important and crucial decisions for the growth of the company. Also to maximize profits and to minimize the costs, a mathematical exploration is needed. The above study is done excluding certain real life factors related to the working of the company such as equity and liabilities, different investments, marketing etc. As an extension to this study, one could also analyze the certain other real life factors mentioned earlier. Another study could be framed to study the increment or reduction in the equity and liabilities of the company over a given period of time. A study could be done on the increase or decrease of income generated from various investment. One could also analyze the marketing strategies of the company and its outcomes in order to obtain useful results which would be required to decide innovative and more efficient marketing strategies.

Bibliography