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Table of content

Rationale

Aim

Background

Profit and cost of goods sold

Differentiation

Integration

Concepts of minima and maxima

Methodology

Data collection and calculation

Calculation of local maximum

Calculation of local minimum

Calculation of local maximum

Conclusion

Reflection

Strength

Weakness

Further scope

Bibliography

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.

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.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

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.

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.

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.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

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.

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.

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

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^{2 }+ 2000000x + 60000000…(i) where y = p (say) denotes the profit and x = t (say) denotes the time.

∴p = - 22820t^{2 }+ 2000000t + 60000000…(i)

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 = - 22820*t*^{2 }+ 2000000*t* + 60000000…(i)

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

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

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

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

Now, we equate the equation obtained to zero.

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

\(\frac{dp}{dt}\) = - 45640*t* + 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 = - 22820t^{2} + 2000000t + 60000000 will have its maximum value at t = 43.82. At t = 43.82, the value of the local maximum is,

*p* = - 22820*t*^{2} + 2000000*t *+ 60000000

*p* = - 22820×43.82^{2 }+ 2000000×43.82 + 60000000

*p* = 103821209.43 (*in INR*)

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

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

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* = 80000000*x*^{2 }+ 600000000*x* + 1000000000…(iii) where y = c (say) denotes the profit and x = t (say) denotes the time.

∴c = 80000000*t*^{2} + 600000000*t* + 1000000000…(iii)

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

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 = 80000000*t*^{2} + 600000000*t* + 1000000000…(*iii*)

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

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

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

Now, we equate the equation obtained to zero.

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

\(\frac{dc}{dt}\) = 160000000*t* + 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 = 80000000*t*^{2} + 600000000*t* + 1000000000 will have its minimum value at t = 3.75.

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

*c* = 80000000*t*^{2} + 600000000*t* + 1000000000

c = 80000000×3.75^{2} + 600000000×3.75 + 1000000000

c = 1337500000(*in INR*)

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

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

\(\frac{dc}{dt}\) = 160000000*t* + 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 = 8000000*t*^{2} + 600000000*t* + 1000000000…(*iii*)

\(\frac{c}{10^8}\) = 0.08*t*^{2} + 6*t* + 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* = - \(\frac{7c}{25000}+\frac{388554725ln(c+497836860000)}{4}\) + *C*_{1}

Where *C*_{1} 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* = - \(\frac{7c}{25000}+\frac{388554725ln(c+497836860000)}{4}\) + *C*_{1}

Putting p = 58780477 and c = 1349071587 we get,

58780477 = - \(\frac{7×1349071587}{25000}+\frac{388554725ln(1349071587+497836860000)}{4}\) + *C*_{1}

*C*_{1} = 58780477 + 377740.04 + 97138681.25×26.936

*C*_{1} = 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*)

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*).

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

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.

- 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.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

- 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.

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.

- “Understanding Cost of Goods Sold – COGS.”
*Investopedia,*https://www.investopedia.com/terms/c/cogs.asp. Accessed 18 Sept. 2021. *Finding Maxima and Minima Using Derivatives.*https://www.mathsisfun.com/calculus/maxima minima.html.%20Accessed%2018%20Sept.%202021.- “Cartesian Coordinate System.”
*Wikipedia*, 30 Aug. 2021.*Wikipedia*, https://en.wikipedia.org/w/index.php title=Cartesian_coordinate_system&oldid=1041463197. - Taken from profit and loss statement of Eastman Auto and Power limited
- Taken from profit and loss statement of Eastman Auto and Power limited

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student