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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.

**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

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

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}*),

\(cov(x, y) =\frac{1}{n}\sum^n_i=1(xi − x̅) (yi − y̅)\), 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,

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

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

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

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,

_{\( rxy= \frac {cov(x, y) }{\sqrt{var(x)}\sqrt {var(y)}}\,where,\)}

_{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,\(cov(x, y)\frac{1}{n}\sum^n_{i=1}(xi − \bar{x}) (yi − \bar{y})\), 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,

_{ }\( rxy =\frac {cov(x, y) }{\sqrt{var(x)}\sqrt {var(y)}}\)

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

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

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*.

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.

**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* = −0.006*x*^{ 2} + 1.8*x* + 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* = −0.006*x* ^{2} + 1.8*x* + 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 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* = −0.006*x*^{ 2} + 1.8*x* + 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}(0.05*x*) + 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}(0.05*x*) + 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.*

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.