Investigating the correlation between number of employees and total profit, marginal profit in Gems and Jewelers, Ahmadabad, India

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.

 

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.

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’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.

 

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

 

\(b_{yx}= \frac{cov(x,y)}{var(x)}\)

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.

 

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 (x_i,y_i), 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.

 

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.

 

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

 

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.

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 2

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 = - 0.006x² + 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):

 

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

 

y = - 0.006x² + 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) is,

 

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

 

= - 2.421 + 1.8 million USD

 

= - 0.621 million USD

 

∴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 = - 0.006x² + 1.8x + 100

 

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

 

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

 

= - 14331.19 million USD

 

∴ Total revenue of the products sold in the year 2010 is - 14331.19 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):

 

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)

 

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

 

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

 

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)

 

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

 

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

= 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) -

 

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)

 

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

 

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

 

Now, \(\frac{d}{dx}\) 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)

 

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

 

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

 

= 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 the 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):

 

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

 

y = x + sin⁡(0.4x)

 

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

 

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

 

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

 

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

 

\(=\frac{2\ cos\ cos\big(\frac{2×205.41}{5}\big)}{5} + 1\) 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 (0.4x)

 

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

 

\(=\frac{225^2}{2}-\frac{5\ cos(\frac{2×255}{5})}{2}-\bigg(\frac{150^2}{2}-\frac{5\ cos\ cos\frac{2×150}{5})}{2}\bigg)\) 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 + 9000csc^-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):

 

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

 

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)

 

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

 

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

 

= 22720.743 million USD

 

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

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 - 0.01x²

 

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

 

Calculation of marginal revenue (in million USD):

 

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

 

y = 3.241 + 3.564x - 0.01x²

 

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

 

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

 

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 - 0.01x²

 

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

 

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

 

= 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) -

 

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

 

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

 

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

 

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

 

= 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):

 

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

 

y = 1.03^x

 

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

 

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

 

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

 

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

 

\(=\frac{(ln\ ln(103)-ln\ ln(100))103^{207.33}}{100^{207.33}}\)

 

≈ 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

 

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

 

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

 

= 59674.02 million USD

 

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

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 = 0.002x² + 150

 

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

 

Calculation of marginal revenue (in million USD) -

 

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

 

y = 0.002x² + 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 = 0.002x² + 150

 

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

 

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

 

= 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 = 4 cos cos (0.2x) + 0.002x² + 150

 

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

 

Calculation of marginal revenue (in million USD):

 

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

 

y = 4 cos cos (0.2x) + 0.002x² + 150

 

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

 

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

 

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 = 4 cos cos (0.2x) + 0.002x² + 150

 

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

 

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

 

= 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 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 = 0.0034x² + 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 = 0.0034x² + 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 × 201.75² + 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 = 0.0034x² + 30

 

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

 

\(=\frac{17×272^3}{15000} + 30 × 272 - \bigg(\frac{17×123^3}{15000}+30×23\bigg) \, \,million \, \,USD\)

 

= 25167.818 million USD

 

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

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,

 

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

 

\(\frac{25}{196.83}+\frac{20}{196.83^2\sqrt{1-\frac{400}{196.83^2}}} \, \, \,million \, \,US\)

 

= 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)

 

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

 

\(= 272\bigg(25\ ln\ ln(272)+sec^{-1}\bigg(\frac{272}{20}\bigg)-25\bigg) -20\, \, ln\bigg(\sqrt{(272-20)(272+20)} + 272 - (125)\)

 

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

 

= 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,

 

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

 

= 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)

 

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

 

_{\(= 110 × 250tan^{-1}\bigg(\frac{250}{100}\bigg)-5500\ ln\ ln(250^2+10000)-(110×120tan^{-1}\bigg(\frac{120}{100}\bigg) - 5500 \, \,ln \, \,ln (1202 +10000)) \, \,million \, \,USD\)}

 

= 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):

 

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

y = 0.5x + sin⁡(0.4x)

 

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

 

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

 

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)

 

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

 

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

 

= 10628.615 million USD

 

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

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 ln (x) + 2500csc^-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):

 

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

 

y = 0.2 ln ln (x) + 2500csc^-1(0.09x)

 

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

 

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

 

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

 

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

 

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

 

= - 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 ln (x) + 2500csc^-1(0.09x)

 

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

 

\(=\frac{1250000\ln \ ln(\sqrt{81×245^2-10000}+9×245)+9×245\ ln(245)+(112500csc^{-1}(\frac{9×245}{100})-9)×245}{45} - \frac{1250000\ln \ ln(\sqrt{81×145^2-10000}+9×145)+9×245\ ln(145)+(112500csc^{-1}(\frac{9×145}{100})-9)×145}{45}​​​​​​​million \,USD\)

 

= 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 - 0.008x²

 

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 = 0.241 + 2.39x - 0.008x²

 

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

 

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

 

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 - 0.008x²

 

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

 

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

 

= 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) -

 

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

 

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 = 20000cot^-1(2x) + 0.3x

 

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

 

 \(= 5000 ln ln (4 × 230^2 + 1) + 20000 × 230 cot^{-1} (2 × 230) +\frac{3×230^2}{20}-\bigg(5000\ ln\ ln(4×120^2+1\bigg) + 20000 × 120cot-1 (2 × 120) + \frac{3×120^2}{20}\bigg)\, million \, \, USD\)

 

= 12280.854 million USD

 

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

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 = 0.0007x² + 100

 

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 = 0.0007x² + 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 = 0.0007x² + 100

 

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

 

\(=\frac{7×255^2}{30000}+100×255-\bigg(\frac{7×160^2}{30000}+100×160\bigg) \, \, million \, \,USD\)

 

= 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 = 0.0012x² + ln 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) -

 

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

 

y = 0.0012x² + ln ln (x) + 80

 

\(\frac{d}{dx}(y)=\frac{d}{dx}(0.0012x2 + ln \, \,ln \,(x) + 80)\)

 

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

 

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 = 0.0012x² + ln ln (x) + 80

 

\(\displaystyle\int\limits^{275}_{\ 130}y=\displaystyle\int\limits^{275}_{\ 130} (0.0012x2 + \, \,ln \, \,ln \,(x) + 80)dx\)

 

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

 

\(=275ln(275)+\frac{275^2}{2500}+79×275-\bigg(130ln(130)+\frac{130^2}{2500}+79×130\bigg) \, \,million \, \,USD\)

 

= 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 cos (0.1x) + 0.0009x² + 100

 

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 = 3 cos cos (0.1x) + 0.0009x² + 100

 

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

 

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

 

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

 

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

 

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

 

= 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 cos (0.1x) + 0.0009x² + 100

 

\(\displaystyle\int\limits^{285}_{\ 125}y=\displaystyle\int\limits^{285}_{\ 125}(3 cos cos (0.1x) + 0.0009x^2 + 10)dx\)

 

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

 

= 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.

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

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) vary 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.

 

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

 

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

 

\(=\frac{1}{n}\sum\limits^{n}_{i=1}x_iy_i-\bar {xy}\)

 

\(Var(x)=\frac{1}{n}\sum\limits_{i}(x_i-\bar x)^2=\frac{1}{n}\sum\limits_{i}x_i^2-\bar x^2\)

 

Similarly,

 

\(Var(y)=\frac{1}{n}\sum\limits_{i}(y_i-\bar y)^2=\frac{1}{n}\sum\limits_{i}y_i^2-\bar y^2\)

 

So, we can write,

 

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

 

\(=\frac{\frac{1}{n}\sum\limits_{i=1}^{n}{x_iy_i-\bar{xy}}}{\sqrt{\frac{1}{n}\sum\limits_{i}{x_{i^2}-x^{-2}}}\sqrt{\frac{1}{n}\sum\limits_{i}{y_{i^2-}y^{-2}}}}\)

 

\(=\frac{n\sum\limits_{i}{x_iy_i-(\sum\limits_iy_i)(\sum\limits_iy_i)}}{\sqrt{n\sum\limits_{i}{x_{i^2}-(\sum\limits_ix_i)^2}}\sqrt{n\sum\limits_{i}{y_{i^2}-(\sum\limits_iy_i)^2}}}\)

 

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

 

\(r_{xy}=\frac{n\sum\limits_{i}{x_iy_i-(\sum\limits_ix_i)(\sum\limits_iy_i)}}{\sqrt{n\sum\limits_{i}{x_{i^2}-(\sum\limits_ix_i)^2}}\sqrt{n\sum\limits_{i}{y_{i^2}-(\sum\limits_iy_i)^2}}}\)

 

\(\frac{10(-19920+1442.1+2131.1+5325+4689.5+4503+2702.7+3436.5+12792)-112150×0.986}{\sqrt{10×628445441-(112150)^2}\sqrt{10×5.98^2-(0.986)^2}}\)

 

= 0.47

 

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

 

\(r_{xy}=\frac{n\sum\limits_{i}{x_iy_i-(\sum\limits_ix_i)(\sum\limits_iy_i)}}{\sqrt{n\sum\limits{x_{i^2}-(\sum\limits_ix_i)^2}}\sqrt{n\sum\limits{y_{i^2}-(\sum\limits_iy_i)^2}}}\)

 

= 0.58

 

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

 

We have,

 

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

 

\(=\frac{n\sum\limits_ix_iy_i-(\sum\limits_{i}x_i)(\sum\limits_iy_i)}{n\sum x_i^2-(\sum\limits_ix_i)^2}\)

 

= 0.46

 

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

 

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

 

\(=\frac{n\sum\limits_ix_iy_i-(\sum\limits_{i}x_i)(\sum\limits_iy_i)}{n\sum x_i^2-(\sum\limits_ix_i)^2}\)

 

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

 

= 0.34

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.

• 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.

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 analysis, 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.

“Regression Analysis.” Wikipedia, 12 July 2021. Wikipedia, https://en.wikipedia.org/w/index.php?title=Regression_analysis&oldid=1033186605. CloseDeleteEdit