I always wondered how come mathematics; the study of numbers and their relationship can be actually useful for a person like me who aspire 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 make useful conclusion.

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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.

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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.

To what extent is there a correlation between the number of employees and the marginal profit, total profit of Gems and Jewelers, Ahmadabad, India. Determine using differentiation of trigonometric function, integration of trig function, regression analysis and Pearson&rsquo;s correlation analysis.

By regression 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.

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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

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\(b_{yx}= \frac{cov(x,y)}{var(x)}\)

By correlation we mean the association or interdependence between two variables. If two variables be 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.

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

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

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

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

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

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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.

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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.

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

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

Raw data for Quantity.svg

Raw data for Revenue.svg

Raw data for cost.svg

Sample calculation for finding the arithmetic mean:

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

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

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Now, we will calculate the marginal revenue and total revenue of the products sold each year from the year 2010 to 2

Variation of revenue.svg

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,

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y = - 0.006x2&nbsp;+ 1.8x + 100

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

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Calculation of marginal revenue (in million USD):

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

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y = - 0.006x2 + 1.8x + 100

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

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

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\(\frac{dy}{dx} = - 0.012x + 1.8\)

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

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

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= - 2.421 + 1.8 million USD

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= - 0.621 million USD

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&there4;Marginal revenue for the year 2010 is - 0.621 million USD

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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.

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y = - 0.006x2 + 1.8x + 100

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\(\displaystyle\int\limits^{272}_{123}y=\displaystyle\int\limits^{272}_{123}(- 0.006x2 + 1.8x + 100)dx\)

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

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= - 14331.19 million USD

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&there4; Total revenue of the products sold in the year 2010 is - 14331.19 million USD.

sold in the year 2011.svg

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,

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y = 22sec-1(0.05x) + 50ln⁡(x)

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

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Calculation of marginal revenue (in million USD):

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

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y = 22sec-1(0.05x) + 50ln⁡(x)

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

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

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

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

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= 0.265 million USD

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&there4;Marginal revenue for the year 2011 is 0.265 million USD.

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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.

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y = 22sec-1(0.05x) + 50ln⁡(x)

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

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\(= (50 × 272 ln (272) + \bigg(22 sec^{-1}\bigg(\frac{272}{20}\bigg)-50\bigg)272-220 \,ln \,ln\bigg(\sqrt{1-\frac{400}{272^2}}+1\bigg))+ 220 \, \,ln \, \,ln \bigg(\sqrt{1-\frac{400}{272^2}}-1\bigg))-\bigg(50×125ln(125)+(22sec^{-1}\bigg(\frac{125}{20}\bigg)-50\bigg)125 - 220 \, \,ln \, \, ln \bigg(\sqrt{1-\frac{400}{125^2}}+1\bigg)+ 220 \, \,ln \, \,ln\bigg(\sqrt{1-\frac{400}{125^2}}-1\bigg)\bigg) \,{million \,USD}\)&nbsp;&nbsp;

= 43449.062 million USD

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&there4;Total revenue of the products sold in the year 2011 is 43449.062 million USD.

sold in the year 2012.svg

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,

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y = 200tan-1(0.009x)

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

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Calculation of marginal revenue (in million USD) -

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

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y = 200tan-1(0.009x)

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

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

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

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

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

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= 0.4181 million USD

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&there4;Marginal revenue for the year 2012 is 0.265 million USD.

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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.

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y = 200tan-1(0.009x)

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

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\(=\frac{200000(\frac{9\times250tan^{-1}\bigg(\frac{9\times250}{1000}\bigg)\frac{In/n(81\times250^2+1000000)}{2}}{1000})}{9}-\frac{200000(\frac{9\times120tan^{-1}\bigg(\frac{9\times120}{1000}\bigg)\frac{In/n(81\times120^2+1000000)}{2}}{1000})}{9}\)million USD

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= 26422.472 million USD

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&there4;Total revenue of the products sold in the year 2012 is 26422.472 million USD.

sold in the year 2013.svg

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

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y = x + sin⁡(0.4x)

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

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Calculation of marginal revenue (in million USD):

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

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y = x + sin⁡(0.4x)

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

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

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

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

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

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= 1.354 million USD

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&there4;Marginal revenue for the year 2013 is 1.354 million USD.

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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.

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y = x + sin (0.4x)

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

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\(=\frac{225^2}{2}-\frac{5\ cos(\frac{2×255}{5})}{2}-\bigg(\frac{150^2}{2}-\frac{5\ cos\ cos\frac{2×150}{5})}{2}\bigg)\)&nbsp;million USD

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= 21259.865 million USD

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&there4;Total revenue of the products sold in the year 2013 is 21259.865 million USD.

sold in the year 2014.svg

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,

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y = 180 + 9000csc-1(x)

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

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Calculation of marginal revenue (in million USD):

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

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y = 180 + 9000csc-1(x)

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

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

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

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

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

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= - 0.228 million USD

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&there4;Marginal revenue for the year 2014 is - 0.228 million USD.

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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.

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y = 180 + 9000csc-1(x)

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

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

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= 22720.743 million USD

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&there4;Total revenue of the products sold in the year 2014 is 22720.743 &nbsp;million US.

sold in the year 2015.svg

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,

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y = 3.241 + 3.564x - 0.01x2

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

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Calculation of marginal revenue (in million USD):

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

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y = 3.241 + 3.564x - 0.01x2

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

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

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

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

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

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= - 0.331 million USD

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&there4;Marginal revenue for the year 2015 is - 0.331 million USD.

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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.

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y = 3.241 + 3.564x - 0.01x2

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

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

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= 40753.935 million USD

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&there4; Total revenue of the products sold in the year 2015 is 40753.935 million USD.