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Table of content

Research question

Rationale

Background information

Key terms used

Regression analysis

Correlation and karl pearson's product moment correlation coefficent

Trigonometric function

Differentiation

Integration

Variables

Independent variables

Dependent variables

Data collection

Analysis

Determination of marginal revenue and total revenue

Determination of marginal cost and total cost

Determination of correlation and regression

Conclusion

Reflection

Strength

Weakness

Further scope

Bibliography

To what extent is there a correlation between the number of employees and the marginal profit, the number of employees and the total profit over the period from the year 2010 to 2019, determined using differentiation, integration of logarithmic, trigonometric, inverse trigonometric, exponential, polynomial functions and regression analysis?

**I always wondered about mathematics;** the study of numbers and their relationship can be actually useful for a person like me who aspires to be a successful entrepreneur someday. Being an inquirer when I explored this, I was exposed to the concept of econometrics which deals with the application of mathematical tools in analyzing financial data and making useful conclusions.

Being involved in the family business an issue that bothered all of us especially during the time of the pandemic was the shortage of the number of workers impacting the financial ratios of the business. Before my father could actually take the decision about cutting down the number of employees I thought of analyzing the financial data of this company for the last ten years using mathematical tools like Differentiation, Integration, and Correlation Analysis to deduce a scientifically supported decision. While preparing the outline I realized that I need to learn certain new formulas and concepts of trigonometric functions for my research.

The study would help prepare the business to take calculated risks and the right decisions to grow it. Also to recover from the loss due to the pandemic situation, the study would be useful.

Dr. Adam Nazha

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

**Revenue of products sold:** Revenue of products sold is the capital earned after selling the quantity of products.

**Cost of products: **The cost of products is the amount of capital spend or the making of the goods to be sold.

**Profit:** Profit the amount of capital left after cutting the amount of capital spend for making the products from the revenue earned by selling them.

**Marginal revenue:** The marginal revenue is the revenue earned from selling one additional product.

**Total revenue:** Total revenue is the revenue earned in total by selling the products, considering the whole year.

**Marginal cost:** The cost added for adding one additional unit of the product.

**Total cost**: Total cost is the total amount of capital used or spend for the making or manufacturing of the products, considering the whole year.

By regression1 of a variable y on another variable x, we mean the dependence of y on x, on the average. In bivariate analysis, one of the major problems is the prediction of the value of the dependent variable y when the value of the independent variable x is known.

Regression coefficient gives the increment in y for a unit increase in x or vice versa. The expression of the regression coefficient (y on x) is given by

b_{yx} =\(\frac {cov(x, y)}{var(x)} \)Where

cov(x, y) denotes the covariance of the two variables x and y

var(x) and var(y) denotes the variances of the variables.

If we are given n pairs of values (x_{i }, y_{i}),i = 1(1)n, of variables x and y then,

cov(x, y) =\(\frac{1}{n}\sum^n_i\)=1(x_{i} − x̅) (y_{i }− y̅), where x̅ and y̅ are the means of the variables x and y respectively

\(cov(x, y) =\frac{1}{n}\displaystyle\sum^{n}_{i=1} x_i \ y_i-\bar x\bar y\)

Var(x) =\(\frac{1}{n}\displaystyle\sum_{i}{(x_i-\bar x)^2}=\frac{1}{n}\sum_{i} x_i2-\bar x^2\)

So, we can write,

b_{yx =}\(\frac{cov(x, y)}{var(x)}\)

\(=\frac{\frac{1}{n}\sum^{n}_{i=1}x_iy_i-\bar x\bar y}{\frac{1}{n}\sum_ix_i^2-\bar x^2}\)

=\(\frac{n\sum_ ix_iy_i-(\sum_ix_i)(\sum_iy_i)}{n\sum_ix_iy_i-(\sum_ix_i)(\sum_iy_i)}\)

By correlation2 we mean the association or interdependence between two variables. If two variables are so related that a change in the magnitude of one of them is accompanied by a change in magnitude of the other, they are said to be correlated.

A measure of the correlation between two variables x and y is Karl Pearson’s product moment correlation coefficient which is defined by,

r_{xy\( = \frac {cov(x, y) }{\sqrt{var(x)}\sqrt {var(y)}}\)where,}

_{cov(x, y)} denotes the covariance of the two variables_{ x} and_{ y; }var_{(x)} and var_{(y) }denotes the variances of the variables.

If we are given n pairs of values (xi , yi),i = 1(1)n, of variables x and y then,cov(x, y)\(\frac{1}{n}\sum^n_{i=1}\)(x_{i} − x̅) (y_{i }− y̅), where x̅ and y̅ are the means of the variables x and y respectively

cov(x, y)=\(\frac{1}{n}\displaystyle \sum^n_{i=1}x_iy_i-\bar x\bar y\)

Var(x) =\(\frac{1}{n} \displaystyle\sum_i{(x_i-\bar x)^2}=\frac{1}{n} \displaystyle\sum_i x_i ^2-\bar x^2\)

Similarly,

Var(y)=\(\frac{1}{n} \displaystyle\sum_i {(y_i-y)^2}=\frac{1}{n}\displaystyle\sum_i y_i2-\bar y^2\)

So, we can write,

r_{xy = }\( \frac {cov(x, y) }{\sqrt{var(x)}\sqrt {var(y)}}\)

=\( \frac {{\frac{1}{n}\sum^n_{i=1}\ x_iy_i-\bar x\bar y} }{\sqrt{\frac{1}{n}\sum _ix_i\ ^2-\bar x^2}\sqrt {\frac{1}{n}\sum_i y i^2-\bar y^2}}\)

\(=\frac{n\sum_ix_iy_i-(\sum_ix_i) (\sum_iy_i)} {\sqrt {n\sum _ix_i2-(\sum_ix_i)^2} \sqrt {n\sum y_i2-(\sum _iy_i)^2 }} \)

Dr. Adam Nazha

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

Trigonometric functions are the functions which relate an angle of a right- angled triangle to the ratios of the length of the two sides. The main functions are sine, cosine and tangent and their reciprocals are cosecant, secant and cotangent respectively.

The sine angle is defined as the ratio of perpendicular upon hypotenuse of a right-angled triangle. The cosine is defined as the ratio of base upon hypotenuse of the right-angled triangle. Tangent is defined as the ratio of perpendicular upon base.

There are also inverse functions of the three main functions sine, cosine and tangent.

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.

\(\frac{∆y}{∆x}=\frac{dy}{∆x}=\frac{f(x + h) − f(x)}{h}\)

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 subintervals, each of length h then,

\(\displaystyle\int^b_a f(x)dx=h\displaystyle \sum^{n-1}_{r=0}f (a+rh)\)

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

Number of employees

**Marginal profit and total profit.**

For a particular year, the values of revenue (R) , quantity of products sold (Q) and cost of products sold (C) were collected as primary data. A graph was plotted with revenue as the y axis and quantity of products sold in the x axis using Desmos. The function generated from the graph was differentiated to obtain a general expression of the marginal revenue. The mean value of the quantity was plugged into the expression obtained to deduce a value of the marginal revenue. The function was integrated taking the minimum value of revenue as the lower limit and the maximum value of the revenue as the upper limit to obtain a numerical value of the total revenue.

A graph was plotted with cost of goods sold in the y axis and quantity of products sold in the x axis using Desmos. The function generated from the graph was differentiated to obtain a general expression of the marginal cost. The mean value of the quantity was plugged into the expression obtained to deduce a value of the marginal cost.

The function was integrated taking the minimum value of revenue as the lower limit and the maximum value of the revenue as the upper limit to obtain a numerical value of the total cost.

The marginal profit was calculated by the formula

Marginal Profit = Marginal Revenue − Marginal Cost

and the total profit was calculated using the formula

Total Profit = Total Revenue − Total Cost

The same process was followed to calculate the total profit and the marginal profit from the year 2010 to the year 2019. Following this, a regression analysis was done to check the extent of correlation between the number of employees and marginal profit; number of employees and the total profit.

Dr. Adam Nazha

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

**Sample calculation for finding the arithmetic mean: **

Arithmetic mean of the quantity of products sold in the year 2010 is,

\(\frac{123 + 144 + 157 + 168 + 177 + 195 + 214 + 226 + 239 + 245 + 261 + 272}{12}=201.75\)

Now, we will calculate the marginal revenue and total revenue of the products sold each year from the year 2010 to 2019.

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2010, we get the equation of the curve as,

y = −0.006x^{ 2} + 1.8x + 100

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2010.

By differentiating and putting the average quantity of products sold for that very year, we will get our marginal revenue for that year.

y = −0.006x ^{2} + 1.8x + 100

\(\frac{d}{dx}(y)=\frac{d}{dx}(−0.006x^ 2 + 1.8x + 100)\)

\(\frac{dy}{dx}\)= -0.006 × 2x + 1.8 + 0

\(\frac{dy}{dx}= −0.012x + 1.8\)

Now, \(\frac{dy}{dx}\) at the point 201.75 (arithmetic mean of the quantity of products sold for the year 2010 as calculated earlier in Table 3) is,

\(\frac{dy}{dx}=\) −0.012 × 201.75 + 1.8

_{= -2.421 + 1.8 million USD}

_{= − 0.621}* million USD*(Here the negative sign indicates a loss in the marginal revenue) ∴Marginal revenue for the year 2010 is −0.621 million USD.

Now, to get the total revenue^{8} of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2010) as the lower and upper limits respectively.

y = −0.006x^{ 2} + 1.8x + 100

\(\displaystyle\int^{272}_{123}y=\displaystyle\int^{272}_{123}(−0.006x^ 2 + 1.8x + 100)dx\)

=\(\bigg(-0.006\frac{ 272^3}{3}+1.8 \frac {272^2}{2}+100 × 272\bigg) -\bigg(-0.006\frac{ 123^3}{3}+1.8 \frac {123^2}{2}+100 × 123\bigg)\)

*= 31343.938 million USD*

∴Total revenue of the products sold in the year 2010 is 31343.938* million USD.*

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2010, we get the equation of the curve as,

y = 22sec^{−1}(0.05x) + 50ln(x)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2011.

By differentiating and putting the average quantity of products sold for that very year, we will get our marginal revenue for that year.

y = 22sec^{−1}(0.05x) + 50ln(x)

\(\frac{d}{dx}(y)=\frac{d}{dx}(22sec^{-1}(0.05x)+\)50 ln ln (x)

\(\frac{dy}{dx}=\frac{400}{x^2\sqrt {1-\frac{400}{x^2}}}+\frac{50}{x}\)

Now, \(\frac{dy}{dx}\)at the point 196.83 (arithmetic mean of the quantity of products sold for the year 2011) is,

\(\frac{dy}{dx}=\frac{440}{196.83^2\sqrt {1-\frac{400}{196.83^2}}}+\frac{50}{196.83}\)

= 0.265 *million USD*

∴Marginal revenue for the year 2011 is 0.265 million USD.

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2011) as the lower and upper limits respectively.

y = 22sec^{−1}(0.05x) + 50ln(x)

\(\displaystyle\int^{272}_{125}y=\displaystyle\int^{272}_{125}(22sec^{−1}(0.05x)+ 50ln (x) )dx \)

\(= (50 × 272ln(272)+\bigg(22sec^{-1}\bigg(\frac{272}{20}\bigg)-50\bigg)272-220\ ln\bigg( \sqrt{1-\frac{400}{272^2}}+1\bigg))+220\ ln\bigg(\sqrt{1-\frac{400}{ 272^2}}-1\bigg))\\-\bigg(50 × 125 ln(125)+\bigg(22sec^{-1}\bigg(\frac{125}{20}\bigg)-50\bigg)125-220\ ln\bigg(\sqrt{1-\frac{400}{ 125^2}}-1\bigg)+220 ln\ \bigg(\sqrt{1-\frac{400}{125^2}}-1\bigg)\bigg)\)_{million usd}

= 43449.062 *million USD*

∴Total revenue of the products sold in the year 2011 is 43449.062* million USD.*

Dr. Adam Nazha

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

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2012, we get the equation of the curve as,

y = 200tan^{−1\((0.009x)\)}

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2012.

By differentiating and putting the average quantity of products sold for that very year, we will get our marginal revenue for that year.

y = 200tan^{−1\((0.009x)\)}

^{\(\frac{d}{dx}(y)=\frac{d}{dx}(200tan^{-1}(0.009x))\)}

^{\(\frac{dy}{dx}=\frac{9}{5(\frac{81x^2}{1000000}+1)}\)}

Now,\(\frac{dy}{dx}\)at the point 202 (arithmetic mean of the quantity of products sold for the year 2012) is,

\(\frac{dy}{dx}=\frac{9}{5(\frac{81x^2}{1000000}+1)}\)

\(\frac{dy}{dx}=\frac{9}{5(\frac{81× 202^2}{1000000}+1)}\)

= 0.4181* million USD*

∴Marginal revenue for the year 2012 is 0.265 *million USD.*

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2012) as the lower and upper limits respectively.

y = 200tan^{−1\((0.009x)\)}

^{\(\displaystyle\int^{250}_{120}y=\displaystyle\int^{250}_{120}(200tan^{−1}(0.009x))dx\)}

=\(\frac{200000(\frac{9×250tan^{-1\big(\frac{9×250}{1000}\big)}}{1000}-\frac{In(81×250^2+1000000)}{2})}{9}\)

-\(\frac{200000(\frac{9×120tan^{-1\big(\frac{9×120}{1000}\big)}}{1000}-\frac{In(81×120^2+1000000)}{2})}{9}\)*million USD*

*= 26422.472 million USD*

∴Total revenue of the products sold in the year 2012 is 26422.472 million USD.

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2013, we get the equation of the curve as,

\(y=x+sin(0.4x)\)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2013.

\(y=x+sin(0.4x)\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(x+sin(0.4x))\)

\(\frac {dy}{dx}=\frac{2cos{\big(\frac{2x}{5}\big)}}{5}+1\)

Now,\(\frac{dy}{dx}\)at the point 205.41 (arithmetic mean of the quantity of products sold for the year 2013) is,

\(\frac {dy}{dx}=\frac{2cos{\big(\frac{2x}{5}\big)}}{5}+1\)

\(=\frac{2cos{\big(\frac{ 2 × 205.41}{5}\big)}}{5}\)+1* **million USD*

= 1.354 *million USD*

∴Marginal revenue for the year 2013 is 1.354 million USD.

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2013) as the lower and upper limits respectively.

\(y=x+sin\ sin(0.4x)\)

\(\displaystyle\int^{255}_{150}y=\displaystyle\int^{255}_{150}(x+sin(0.4x))dx\)

\(=\frac{255^2}{2}-\frac{5cos(\frac{2×255}{5})}{2}-\bigg(\frac{150^2}{2}-\frac{5cos\ cos(\frac{2×150}{5})}{2}\bigg)\)*million USD*

= 21259.865* million USD*

∴Total revenue of the products sold in the year 2013 is 21259.865 *million USD.*

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2014, we get the equation of the curve as,

\(Y=180+9000\csc^{-1}(x)\)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2014.

\(y=180+9000csc^{-1}(x)\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(180 + 9000csc^{−1}(x))\)

\(=-\frac{9000}{x^2\sqrt {1-\frac{1}{x^2}}}\)

Now, \(\frac{dy}{dx}\) at the point 198.67 (arithmetic mean of the quantity of products sold for the year 2014) is,

\(\frac{dy}{dx}=-\frac{9000}{x^2\sqrt {1-\frac{1}{x^2}}}\)

\(=-\frac{9000}{198.67^2\sqrt {1-\frac{1}{198.67^2}}}\)*million USD*

= −0.228 *million USD*

∴Marginal revenue for the year 2014 is −0.228 *million USD.*

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2014) as the lower and upper limits respectively.

\(y=180+9000csc^{-1}(x)\)

\(\displaystyle\int^{245}_{145}y=\displaystyle\int^{245}_{145}(180+9000csc^{-1}(x))dx\)

\(=9000(245csc^{-1}(245)+\frac{ln\bigg({\sqrt {1-\frac{1}{245^2}+1}\bigg)}}{2}-\frac{ln\bigg({\sqrt {1-\frac{1}{245^2}-1}\bigg)}}{2})+180× 245\\\ -(9000\bigg(145csc^{-1}(145)+\frac{ln\bigg({\sqrt {1-\frac{1}{145^2}+1}\bigg)}}{2}-\frac{ln\bigg({\sqrt {1-\frac{1}{145^2}-1}\bigg)}}{2})+(180× 145\bigg)\)*million USD*

= 22720.743* million USD*

∴Total revenue of the products sold in the year 2014 is 22720.743 million USD.

Dr. Adam Nazha

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

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2015, we get the equation of the curve as,

\(y=3.241+3.564x-0.01x^2\)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2015.

\(y=3.241+3.564x-0.01x^2\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(3.241+3.564x-0.01x^2)\)

\(=\frac{891}{250}-\frac{x}{50}\)

Now, \(\frac{dy}{dx}\) at the point 194.75 (arithmetic mean of the quantity of products sold for the year 2015) is,

\(\frac{dy}{dx}=\frac{891}{250}-\frac{x}{50}\)

\(=\frac{891}{250}-\frac{194.75}{50}\)*million USD*

= −0.331* million USD*

∴Marginal revenue for the year 2015 is −0.331 *million USD.*

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2015) as the lower and upper limits respectively.

\(y=3.241+3.564x-0.01x^2\)

\(\displaystyle\int^{265}_{130}y=\displaystyle\int^{265}_{130}(3.241+3.564x-0.01x^2)dx\)

\(=-\frac{265(10 × 265^2 − 5346 × 265 − 9723)}{3000}-\bigg(\frac{130(10 × 130^2 − 5346 × 130 − 9723) }{3000}\bigg)\)*million USD*

*= 40753.935 million USD*

∴Total revenue of the products sold in the year 2015 is 40753.935* million** USD.*

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2016, we get the equation of the curve as,

\(y=19000cot^{-1}(x)+0.8x\)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2016.

\(y=19000cot^{-1}(x)+0.8x\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(19000cot^{-1}(x)+0.8x\)

\(\frac{dy}{dx}=\frac{4}{5}-\frac{19000}{x^2+1}\)

Now,\(\frac{dy}{dx}\)at the point 184.17 (arithmetic mean of the quantity of products sold for the year 2016) is,

\(\frac{dy}{dx}=19000cot^{-1}(x)+0.8x\)

\(=19000cot^{-1}(184.17)+0.8× 184.17\)* million USD*

=0.239 million USD

∴Marginal revenue for the year 2016 is 0.239 million USD.

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2016) as the lower and upper limits respectively.

\(Y=19000cot^{-1}(x)+0.8x\)

\(\displaystyle\int^{230}_{120}y=\displaystyle\int^{230}_{120}(19000cot^{-1}(x)+0.8x)dx\)

\(=\frac{2(23750ln (230^2 + 1) + 230(47500cot^{−1}(230) + 230))}{5}\\ -\frac{2(23750ln (120^2 + 1) -120(47500cot^{−1}(120) +120))}{5}\)*million USD*

= 27761.003* million USD*

∴Total revenue of the products sold in the year 2016 is 27761.003 *million USD.*

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2017, we get the equation of the curve as,

\(y=1.03^x\)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2017.

\(y=1.03^x\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(1.03^x)\)

\(=\frac{(ln\ ln(103)-ln(100)\ )103^x}{100^x}\)

Now, \(\frac{dy}{dx}\)at the point 207.33 (arithmetic mean of the quantity of products sold for the year 2017) is,

\(\frac{dy}{dx}=\frac{(ln (103) − ln (100) \ )103^x}{100^x}\)

\(=\frac{(ln (103) − ln (100) \ )103^{207.33}}{100^{207.33}}\)million USD

*≈ 0 million USD*

∴Marginal revenue for the year 2017 is 0 million USD.

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2017) as the lower and upper limits respectively.

\(y=1.03^x\)

\(\displaystyle\int^{255}_{160}y=\displaystyle\int^{255}_{160}(1.03^x)dx\)

\(=\frac{103^{255}}{(ln (103) − ln (100)\ )100^{255}}-\frac{103^{160}}{(ln (103) − ln (100)\ )100^{160}}\)*million USD*

= 59674.02 *million USD*

∴Total revenue of the products sold in the year 2017 is 59674.02* **million USD.*

Dr. Adam Nazha

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

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2018, we get the equation of the curve as,

\(Y=0.002x^2+150\)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2018.

\(y=0.002x^2+150\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.002x^ 2 + 150)\)

\(\frac{dy}{dx}=\frac{x}{250}\)

Now,\(\frac{dy}{dx}\) at the point 201.25 (arithmetic mean of the quantity of products sold for the year 2018) is,

\(\frac{dy}{dx}=\frac{x}{250}\)

\(=\frac{201.25}{250}\)*million USD*

= 0.805 *million USD*

∴Marginal revenue for the year 2018 is 0.805 million USD.

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2018) as the lower and upper limits respectively.

\(y=0.002x^2+150\)

\(\displaystyle\int^{275}_{130}y=\displaystyle\int^{275}_{130}(0.002x^ 2 + 150)dx\)

\(=\frac{275^3}{1500}+150× 275-\bigg(\frac{130^3}{1500}+150× 130\bigg)\)*million USD*

= 34149.917 *million USD*

∴Total revenue of the products sold in the year 2017 is 34149.917* **million USD.*

From the above graph, by plotting the values of quantity of products sold and revenue of the products sold for the year 2019, we get the equation of the curve as,

\(y=4cos(0.2x)+0.002x^2+150\)

Where, y = the revenue of the products sold and x = the quantity of the products sold.

**Calculation of marginal revenue (in million USD):**

We will now differentiate the equation obtained to get a general expression of the marginal revenue for the year 2019.

\(y=4cos(0.2x)+0.002x^2+150\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(4cos (0.2x) + 0.002x^ 2 + 150)\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(4cos (0.2x))+\frac{d}{dx}(0.002x^ 2)+\frac{d}{dx}(150)\)

\(\frac{dy}{dx}=4(-sin(0.2x))\bigg(\frac{d}{dx}(0.2x)\bigg)+2× 0.002x^ {2−1} + 0\)

\(\frac{dy}{dx}=-4× 0.2 × sin(0.2x) + 0.004x\)

\(\frac{dy}{dx}=−0.8 × sin(0.2x) + 0.004x\)

Now, \(\frac{dy}{dx}\) at the point 206.25 (arithmetic mean of the quantity of products sold for the year 2019) is,

\(\frac{dy}{dx}=\frac{x}{250}-\frac{4sin(\frac{x}{5})}{5}\)

\(=\frac{206.25}{250}-\frac{4sin(\frac{206.25}{5})}{5}\)*million USD = 1.143 million USD*

∴Marginal revenue for the year 2019 is 1.143 *million USD.*

Now, to get the total revenue of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2019) as the lower and upper limits respectively.

\(y=4cos(0.2x)+0.002x^2+150\)

\(\displaystyle\int^{285}_{125}y=\displaystyle\int^{285}_{125}(4cos(0.2x) + 0.002x ^2 +150)dx\)

\(=\frac{285^3}{1500}+150× 285 + 20sin\bigg(\frac{285}{5}\bigg)-\bigg(\frac{125^3}{1500}+ 150 × 125 + 20sin(\frac{125}{5})\bigg)\)*million USD*

= 38142.037 *million USD*

∴Total revenue of the products sold in the year 2019 is 38142.037 *million USD.*

We calculated the marginal revenue and the total revenue of the products sold each year from the year 2010 to 2019.

Now, we will calculate the marginal cost and the total cost^{17 }of the products sold each year from the year 2010 to 2019.

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2010, we get the equation of the curve as,

\(y = 0.0034x^ 2 + 30\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

By differentiating and putting the average quantity of products sold for that very year, we will get our marginal cost for that year.

\(y = 0.0034x^ 2 + 30\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.0034x^ 2 + 30)\)

\(\frac{dy}{dx}=\frac{17x}{2500}\)

Now, \(\frac{dy}{dx}\)at the point 201.75 (arithmetic mean of the quantity of products sold for the year 2010) is,

\(\frac{dy}{dx}=0.0034x^ 2 + 30\)

\(= 0.0034 × 201.75^2 + 30\) million USD

= 1.371 *million **USD*

∴Marginal cost for the year 2010 is 1.371 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2010) as the lower and upper limits respectively.

\(y = 0.0034x^ 2 + 30\)

\(\displaystyle\int^{272}_{123}y=\displaystyle\int^{272}_{123}(0.0034x^ 2 + 30)dx\)

\(=\frac{17 × 272^3}{15000}+ 30 × 272 −\bigg(\frac{17 × 123^3}{15000}+ 30 × 123\bigg)\)million USD

= 25167.818 *million **USD*

∴Total cost of the products sold in the year 2010 is 25167.818 million USD.

Dr. Adam Nazha

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

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2011, we get the equation of the curve as,

\(y = sec^{−1}(0.05x) + 25ln(x)\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

By differentiating and putting the average quantity of products sold for that very year, we will get our marginal cost for that year.

\(y = sec^{−1}(0.05x) + 25ln(x)\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(sec^{−1}(0.05x) + 25ln(x))\)

\(\frac{dy}{dx}=\frac{25}{x}+ \frac {20 }{x^2\sqrt {1-\frac{400}{x^2}}}\)

Now, \(\frac{dy}{dx}\) at the point 196.83 (arithmetic mean of the quantity of products sold for the year 2011) is,

\(\frac{dy}{dx}=\frac{25}{x}+\frac{20}{x^2 \sqrt {1-\frac{400}{x^2}}}\)

\(=\frac{25}{196.83}+ \frac {20 }{196.83^2\sqrt {1-\frac{400}{196.83^2}}}\)*million USD*

= 0.127* million USD*

∴Marginal cost for the year 2011 is 0.127 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2011) as the lower and upper limits respectively.

\(y=sec^{-1}(0.05x)+25ln(x)\)

\(\displaystyle\int^{272}_{125}y=\displaystyle\int^{272}_{125}(sec^{−1}(0.05x) + 25ln(x))dx\)

\(=272\bigg(25\ ln(272) + sec^{−1}\bigg(\frac{272}{20}\bigg)-25\bigg)-20ln(\sqrt{(272-20)(272+20)}+272 \)

\(-(125\bigg(25\ ln(125)+sec^{-1}\bigg(\frac{125}{20}\bigg)-25\bigg)\)

\(-20\ ln\bigg(\sqrt{(125-20)(125+20)}+125\bigg)\)*million USD*

= 19571.297* *million *USD*

∴Total cost of the products sold in the year 2011 is 19571.297 *million USD.*

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2012, we get the equation of the curve as,

\(Y=110tan^{-1}(0.01x)\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

By differentiating and putting the average quantity of products sold for that very year, we will get our marginal cost for that year.

\(y = 110tan^{−1} (0.01x)\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(110tan^{-1} (0.01x))\)

\(\frac{dy}{dx}=\frac{11}{10(\frac{x^2}{10000}+1)}\)

Now,\(\frac{dy}{dx}\)at the point 202 (arithmetic mean of the quantity of products sold for the year 2012) is,

\(\frac{dy}{dx}=\frac{11}{10(\frac{x^2}{10000}+1)}\)

\(=\frac{11}{10(\frac{202^2}{10000}+1)}\)*million USD*

= 0.216 *million USD*

∴Marginal cost for the year 2012 is 0.216 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2012) as the lower and upper limits respectively.

\(y = 110tan^{−1} (0.01x)\)

\(\displaystyle\int^{250}_{120}y=\displaystyle\int^{250}_{120}(110tan^{−1} (0.01x))dx\)

\(=110× 250tan^{-1}\bigg(\frac{250}{100}\bigg)-5500\ ln(250^2+10000)\)

\(-\bigg(110× 120tan^{-1}\bigg(\frac{120}{100}\bigg)− 5500 \ ln(120^2 + 10000)\bigg)\)million *USD*

= 15179.488* *million* USD*

∴Total cost of the products sold in the year 2012 is 15179.488 million USD.

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2013, we get the equation of the curve as,

\(y = 0.5x + sin(0.4x)\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

\(y = 0.5x + sin(0.4x)\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.5x + sin(0.4x))\)

\(\frac{dy}{dx}=\frac{2cos(\frac{2x}{5})}{5}+\frac{1}{2}\)

Now, \(\frac{dy}{dx}\) at the point 205.41 (arithmetic mean of the quantity of products sold for the year 2013) is,

\(\frac{dy}{dx}=\frac{2cos(\frac{2x}{5})}{5}+\frac{1}{2}\)

\(=\frac{2cos(\frac{2× 205.41}{5})}{5}+\frac{1}{2}\)*million USD*

= 0.854 *million USD*

∴Marginal cost for the year 2013 is 0.854 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2013) as the lower and upper limits respectively.

\(y = 0.5x + sin(0.4x)\)

\(\displaystyle\int^{255}_{150}y=\displaystyle\int^{255}_{150}(0.5x + sin(0.4x))dx\)

\(=\frac{255^2-10cos(\frac{2× 255}{5})}{4}-\bigg(\frac{150^2-10\ cos(\frac{2× 150}{5})}{4}\bigg)\)*million USD*

= 10628.615* million USD*

∴Total cost of the products sold in the year 2013 is 10628.615 million USD.

Dr. Adam Nazha

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

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2014, we get the equation of the curve as,

\(y = 0.2 ln(x) + 2500\ csc^{−1} (0.09x)\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

\(y = 0.2 \ ln(x) + 2500 csc^{−1} (0.09x)\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.2\ ln(x) + 2500 \csc^{−1} (0.09x))\)

\(\frac{dy}{dx}=\frac{1}{5x}+ \frac {250000 }{9x^2\sqrt {1-\frac{10000}{81x^2}}}\)

Now, \(\frac{dy}{dx}\) at the point 198.67 (arithmetic mean of the quantity of products sold for the year 2014) is,

\(\frac{dy}{dx}=\frac{2cos(\frac{2x}{5})}{5}+\frac{1}{2}\)

\(=\frac{2cos(\frac{2 × 198.67}{5})}{5}+\frac{1}{2}\)*million USD*

= −0.703* million USD*

∴Marginal cost for the year 2014 is -0.703 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2014) as the lower and upper limits respectively.

\(y = 0.2\ ln(x) + 2500csc^{−1} (0.09x)\)

\(\displaystyle\int^{245}_{145}y=\displaystyle\int^{245}_{145}(0.2 \ ln(x) + 2500csc^{−1} (0.09x))dx\)

\(=\frac{1250000\ ln(\sqrt {81× 245^2-10000}+9× 245)+9× 245ln(245)+(112500csc^{-1}(\frac{9× 245}{100})-9)× 245}{45}\)

\(=\frac{1250000\ ln(\sqrt {81× 145^2-10000}+9× 145)+9× 145ln(145)+(112500csc^{-1}(\frac{9× 145}{100})-9)× 145}{45}\)*million USD*

= 14684.207 *million* *USD*

∴Total cost of the products sold in the year 2014 is 14684.207* million USD.*

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2015, we get the equation of the curve as,

\(y=0.241+2.39x-0.008x^2\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

\(y=0.241+2.39x-0.008x^2\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.241+2.39x-0.008x^2)\)

\(\frac{dy}{dx}=\frac{239}{100}-\frac{2x}{125}\)

Now, \(\frac{dy}{dx}\) at the point 194.75 (arithmetic mean of the quantity of products sold for the year 2015) is,

\(\frac{dy}{dx}=\frac{239}{100}-\frac{2x}{125}\)

\(=\frac{239}{100}-\frac{2× 194.75}{125}\)*million USD*

= −0.726* million USD*

∴Marginal cost for the year 2015 is -0.726 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2015) as the lower and upper limits respectively.

\(y=0.241+2.39x-0.008x^2\)

\(\displaystyle\int^{265}_{130}y=\displaystyle\int^{265}_{130}(0.241\ +2.39x-0.008x^2)dx\)

\(=-\frac{265(8 × 265^2 − 3585 × 265 − 723)}{3000}-\bigg(-\frac{130(8 × 130^2 − 3585 × 130 − 723)}{3000 }\bigg)\)million USD

= 19988.91* **million** USD*

∴Total cost of the products sold in the year 2015 is 14684.207 *million USD.*

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2016, we get the equation of the curve as,

\(y=20000cot^{-1}(2x)+0.3x\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

\(y=20000cot^{-1}(2x)\ +0.3x\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(20000cot^{−1}(2x) + 0.3x)\)

\(\frac{dy}{dx}=\frac {3}{10}-\frac{40000}{4x^2+1}\)

Now, \(\frac{dy}{dx}\) at the point 184.17 (arithmetic mean of the quantity of products sold for the year 2016) is,

\(\frac{dy}{dx}=\frac{3}{10}-\frac{40000}{4x^2+1}\)

\(=\frac{3}{10}\frac{40000}{4× 184.17^2+1}\)*million USD*

= 0.005 million USD

∴Marginal cost for the year 2016 is 0.005 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2016) as the lower and upper limits respectively.

\(y=20000\ cot^{-1}(2x)+0.3x\)

\(\displaystyle\int^{230}_{120}y=\displaystyle\int^{230}_{120}(20000 cot^{−1}(2x) + 0.3x)dx\)

\(=5000\ ln(4× 230^2+1)+20000× 230 \ cot^{−1}(2 × 230)+\frac{3 × 230^2}{20}\)

\(-\bigg(5000\ ln(4× 120^2+1)+20000 × 120\ cot^{−1}(2 × 120)\ + \frac{3× 120^2}{20}\bigg)\)*million USD*

= 12280.854 *million **USD*

∴Total cost of the products sold in the year 2016 is 14684.207 *million USD.*

Dr. Adam Nazha

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

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2017, we get the equation of the curve as,

\(y=0.0007x^2+100\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

\(y=0.0007x^2 +100\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.0007x^2+100)\)

\(\frac{dy}{dx}=\frac{7x}{5000}\)

Now, \(\frac{dy}{dx}\)at the point 207.33 (arithmetic mean of the quantity of products sold for the year 2017) is,

\(\frac{dy}{dx}=\frac{7x}{5000}\)

\(=\frac{7 × 207.33}{5000}\)*million USD*

= 0.290 *million USD*

∴Marginal cost for the year 2017 is 0.290 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2017) as the lower and upper limits respectively.

\(y = 0.0007x^ 2 + 100\)

\(\displaystyle\int^{255}_{160}y=\displaystyle\int^{255}_{160}(0.0007x^ 2 + 100)dx\)

\(=\frac{7 × 255^3}{30000}+ 100 × 255 −\bigg(\frac{7 × 160^3}{30000}+ 100 × 160\bigg)\)*million USD*

= 12413.254 million USD

∴Total cost of the products sold in the year 2017 is 12413.254* million USD.*

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2018, we get the equation of the curve as,

\(y = 0.0012x^ 2 + ln(x) + 80\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

\(y = 0.0012x^ 2 + ln(x) + 80\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.0012x^ 2 + ln(x) + 80)\)

\(\frac{dy}{dx}=\frac{3x}{1250}+\frac{1}{x}\)

Now, \(\frac{dy}{dx}\)at the point 201.25 (arithmetic mean of the quantity of products sold for the year 2018) is,

\(\frac{dy}{dx}=\frac{3x}{1250}+\frac{1}{x}\)

\(=\frac{3 × 201.25}{1250}+\frac{1}{201.25}\)*million USD*

= 0.487 *million** USD*

∴Marginal cost for the year 2018 is 0.487 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2018) as the lower and upper limits respectively.

\(y = 0.0012x^ 2 + ln(x) + 80\)

\(\displaystyle\int^{275}_{130}y=\displaystyle\int^{275}_{130}(0.0012x^ 2 + ln(x) + 80)dx\)

\(=\displaystyle\int^{275}_{130}0.0012x^ 2dx+\displaystyle\int^{275}_{130}ln(x) dx+\displaystyle\int^{275}_{130}80dx\)

\(= 275ln(275)+\frac{275^3}{2500}+ 79 × 275 −\bigg(130ln(130)+\frac{130^3}{2500}+ 79 × 130\bigg)\)*million USD*

= 19806.782 *million USD*

∴Total cost of the products sold in the year 2018 is 19806.782 *million USD.*

From the above graph, by plotting the values of quantity of products sold and cost of the products sold for the year 2019, we get the equation of the curve as,

\(y = 3 \ cos(0.1x) + 0.0009x^ 2 + 100\)

Where, y = the cost of the products sold and x = the quantity of the products sold.

**Calculation of marginal cost (in million USD):**

\(y = 3\ cos(0.1x) + 0.0009x^ 2 + 100\)

\(\frac{d}{dx}(y)=\frac{d}{dx}(3 \ cos(0.1x) + 0.0009x^ 2 + 100)\)

\(\frac{dy}{dx}=\frac{9x}{5000}-\frac{3sin(\frac{x}{10})}{10}\)

Now,\(\frac{dy}{dx}\)at the point 206.25 (arithmetic mean of the quantity of products sold for the year 2019) is,

\(\frac{dy}{dx}=\frac{9x}{5000}-\frac{3sin(\frac{x}{10})}{10}\)

\(=\frac{9 × 206.25}{5000}-\frac{3sin(206.25)}{10}\)*million USD*

*= 0.077 million USD*

∴Marginal cost for the year 2019 is 0.077 million USD.

Now, to get the total cost of the products sold, we have to integrate the function taking the least and the most quantity of products sold (for the year 2019) as the lower and upper limits respectively.

\(y = 3 cos(0.1x) + 0.0009x^ 2 + 100\)

\(\displaystyle\int^{285}_{125}y=\displaystyle\int^{285}_{125}(3 \ cos(0.1x) + 0.0009x ^2 + 100)dx\)

\(=\frac{1000009 × 285}{10000}+30\ sin\bigg(\frac{285}{10}\bigg)-\bigg(\frac{1000009 × 125}{10000}+30\ sin\bigg(\frac{125}{10}\bigg)\bigg)\)*million USD*

*= 15995.420 million USD*

∴Total cost of the products sold in the year 2018 is 19806.782 million USD.

Now, we will calculate the marginal profit and total profit from marginal revenue, total revenue, marginal cost and total cost.

Dr. Adam Nazha

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

Year

Marginal

Total Revenue (TR)

Marginal Cost (MC)

Total Cost (TC)

Marginal Profit (MP)

Total Profit (TP)

2010

-0.621

31343.938

1.371

25167.818

-1.992

-39499.008

2011

0.265

43449.062

0.127

19571.297

0.138

23877.765

2012

0.418

26422.472

0.216

15179.488

0.202

11242.984

2013

1.354

21259.865

0.854

10628.615

0.5

10631.25

2014

-0.288

22720.743

-0.703

14684.207

0.415

8036.536

2015

-0.331

40753.935

-0.726

19988.91

0.395

20765.025

2016

0.239

27761.003

0.005

12280.854

0.234

15480.149

2017

0

59674.02

0.290

12413.254

-0.290

47260.766

2018

0.805

34149.917

0.487

19806.782

0.318

14343.135

2019

1.143

38142.037

0.077

19806.782

1.066

18335.255

**Sample calculation:**

Marginal profit and total profit for the year 2010 are,

Marginal revenue – Marginal cost = Marginal Profit = (-0.621 - (-1.371)) million USD = - 1.992 million USD

Total revenue – Total cost = Total Profit = (-14331.19 - 25167.818) million USD = - 39499.008 million USD

Year

No. of Employees (N)

Marginal Profit (MP)

Total Profit (TP)

2010

10000

-1.992

6176.12

2011

10450

0.138

23877.765

2012

10550

0.202

11242.984

2013

10650

0.5

10631.25

2014

11300

0.415

80366.536

2015

11400

0.395

20765.025

2016

11550

0.234

15480.149

2017

11850

-0.290

47260.766

2018

12000

0.318

14343.135

2019

12400

1.066

18335.255

Now, we will find the correlation between the number of employees and marginal profit, number of employees and total profit.

We will find out the correlation using Karl Pearson’s product moment correlation coefficient which is given by,

\(r_{xy} \frac {cov(x, y) }{\sqrt{var(x)}\sqrt {var(y)}}\)where,

cov(x, y) denotes the covariance of the two variables x and y; var(x) and var(y) denotes the variances of the variables.

If we are given n pairs of values (xi , yi),i = 1(1)n, of variables x and y.

Here, x is the number of employees and y is the marginal profit.

cov(x, y) = \(\frac{1}{n}\sum^n_{i=1}(x_i-\bar x)(y_i-\bar y)\) where x̅and y̅ are the means of the variables x and y

respectively

\(=\frac{1}{n}\displaystyle\sum^{n}_{i=1} x_i \ y_i-\bar x\bar y\)

\(var(x)=\frac{1}{n}\displaystyle\sum_{1}(x_i-\bar x)^2=\frac{1}{n}\displaystyle\sum_{i} x_i^\ {2}-\bar x^2\)

Similarly,

\(var(y)=\frac{1}{n}\displaystyle\sum _{i}(y_i-y)^2=\frac{1}{n}\displaystyle\sum_{i}y_i^\ 2\ -\bar y^2\)

So, we can write,

\(r_{xy}=\frac {cov(x, y) }{\sqrt{var(x)}\sqrt {var(y)}}\)

\(=\frac{\frac{1}{n}\sum^n_{i=1\ x_iy_i-\bar x\bar y}}{\sqrt{\frac{1}{n}\sum_ix_i^2-\bar x^2}\sqrt{\frac{1}{n}\sum_iy_i^2-\bar y^2}}\)

\(\)

\(=\frac{n\sum_{i\ x_iy_i-(\sum_ix_i)(\sum_iy_i)}}{\sqrt{n\sum x_i^2-(\sum_ix_i)^2 }\sqrt{{n}\sum y_i^2-(\sum_iy_i)^2}}\)

Now, we will find the correlation between number of employees and marginal profit.

\(r_{xy}=\frac{n\sum_{i\ x_iy_i-(\sum_ix_i)(\sum_iy_i)}}{\sqrt{n\sum x_i^2-(\sum_ix_i)^2 }\sqrt{{n}\sum y_i^2-(\sum_iy_i)^2}}\)

\(=\frac{{10(−19920 + 1442.1 + 2131.1 + 5325 + 4689.5 + 4503 + 2702.7 + 3436.5 + 12792) − 112150 × 0.986)}}{\sqrt{10 × 628445441 − (112150)^2 }\sqrt{10 × 5.982^2 − (0.986)^2}}\)

= 0.47

Now, we will find the correlation between number of employees and total profit.

\(r_{xy}=\frac{n\sum_{i\ x_iy_i-(\sum_ix_i)(\sum_iy_i)}}{\sqrt{n\sum x_i^2-(\sum_ix_i)^2 }\sqrt{{n}\sum y_i^2-(\sum_iy_i)^2}}\)

= 0.58

Now, we will find out the regression coefficient between the number of employees and total profit.

We have,

\(b_{yx}=\frac{cov(x, y)}{var(x)}\)

\(=\frac{n\sum_ix_iy_i-(\sum_ix_i)(\sum_iy_i)}{n\sum_ix_iy_i-(\sum_ix_i)(\sum_iy_i)}\)

= 0.46

Now, we will find out the regression coefficient between the number of employees and marginal profit.

We have,

\(b_{yx}=\frac{cov(x, y)}{var(x)}\)

\(=\frac{n\sum_ix_iy_i-(\sum_ix_i)(\sum_iy_i)}{n\sum_ix_iy_i-(\sum_ix_i)(\sum_iy_i)}\)

\(=\frac{10(−19920 + 1442.1 + 2131.1 + 5325 + 4689.5 + 4503 + 2702.7 + 3436.5 + 12792) − 112150 × 0.986}{10 × 628445441 − (112150)^2}\)

= 0.34

To what extent is there a correlation between the number of employees and the marginal profit, the number of employees and the total profit over the period from the year 2010 to 2019, determined using differentiation, integration of logarithmic, trigonometric, inverse trigonometric, exponential, polynomial functions and regression analysis?

At first, we took a data set of values of quantity of products sold, revenue of products sold for each year, starting from the year 2010 to the year 2019. Then, we analyzed the trend between the quantity of products sold and the revenue of the products sold and following the trend, we defined the revenue of products sold as a function of the quantity of products sold and from there, we calculated the marginal revenue and the total revenue. Similarly, we took a data set of values of the cost of products for each year, from the year 2010 to 2019 and calculated the marginal cost and total cost by analyzing the trend between them. Thereafter, we found out the marginal profit and total profit and took the data set of number of employees and found out the correlation and regression between them. From the correlation and regression coefficient we can conclude the following:

- The correlation coefficient between the number of employees and marginal profit is 0.47, which is a positive correlation, which means an increment in the number of employees leads to an increment in the marginal profit. However, there is not a strong correlation between them.
- The correlation between the number of employees and total profit is 0.58, which is a positive correlation, which means an increment in the number of employees leads to an increment in the total profit. The correlation is stronger than that between the number of employees and marginal profit.
- The regression coefficient between the number of employees and marginal profit is 0.34, which means by a unit increase in the number of employees, the increment in the marginal profit is 0.34.
- The regression coefficient between the number of employees and total profit is 0.46, which means by a unit increase in the number of employees, the increment in the total profit is 0.46.
- Therefore, we can conclude that the increment in the workforce of the company may lead to increment in the costs but leads to an even greater increment in the overall total profit.

Dr. Adam Nazha

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

- A pretty wide range of data of ten years has been used which makes the mathematical analysis more relevant to use in real life. The data set of values has been analyzed using graphs and studying the trend, the values and the correlation and regression between them have been mathematically analyzed.
- The total profit and marginal profit has been calculated from for each year from the total revenue, total cost, marginal revenue and marginal cost calculated for each year which makes the analysis even more coherent as the data has been minutely examined and processed.
- The data has been collected from an authentic source, which makes the analysis

and the discussion relevant to real life situations.

- The analysis and the discussions made have not considered some other factors related to a real business which makes the study less coherent. The costs and the profit may vary to a certain extent if other factors like employee’s insurance, tax, bills etc are included.
- The study may suggest what changes are needed to make a business model efficient but again; not considering certain realistic factors along with the recently encountered pandemic situation may vary the study to a great extent.
- The study could be more efficient if the sample size was bigger. Also the correlation and regression analysis would have been more appropriate if higher level statistical and mathematical tools were used.

Dr. Adam Nazha

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

The data used to do the study the trends between the quantity of products sold, revenue and cost is not very wide ranged. The study would have been more coherent if a larger sample size was used. The mathematical analysis would have been more efficient if realistic factors related to a business and higher level statistical and mathematical tools like time series analysis29, sampling theory, non linear regression analysis, double integration etc were used. The trends between the quantity of products, revenue of products sold and cost of products could be more efficiently analyzed and could be a more perfect fit to the data by having a detailed knowledge on mathematical functions.

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