Mathematics AI SL
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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

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Research question

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?

Rationale

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.

Background information

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  • Key terms used

    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.

    Regression analysis

    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 (xi , yi),i = 1(1)n, of variables x and y then,

     

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

     

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

     

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

     

    Correlation and karl pearson's product moment correlation coefficent

    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 − x̅) (yi − y̅), where x̅ and y̅ are the means of the variables x and y respectively

     

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

     

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

     

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

     

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

     

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  • Trigonometric function

    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.

    Differentiation

    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.

    Integration

    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.

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  • Variables

    Independent variables

    Number of employees

    Dependent variables

    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.

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  • Data collection

    Figure 1 - Table On Raw Data For Quantity Of Products Sold From The Year 2010-2019
    Figure 2 - Table On Raw Data For Revenue Of Product Sold From The Year 2010-2019 (In Million USD)
    Figure 3 - Table On Raw Data For Cost Of Goods Sold (COGS) From The Year 2010-2019

    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.

    Analysis

    Determination of marginal revenue and total revenue

    Figure 4 - Variation Of Revenue Of Products Sold In The Year 2010

    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 revenue8 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.

    Figure 5 - Variation Of Revenue Of Products Sold In The Year 2011

    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.

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  • Figure 6 - Variation Of Revenue Of Products Sold In The Year 2012

    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 = 200tan1\((0.009x)\)

     

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

     

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

     

    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.

    Figure 7 -Variation Of Revenue Of Products Sold In The Year 2013

    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.

     

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

     

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

     

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

     

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

     

    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.

    Figure 8 - Variation Of Revenue Of Products Sold In The Year 2014

    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.

     

    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.

     

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

     

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

     

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

     

    = 22720.743 million USD

     

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

     

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  • Figure 9 - Variation Of Revenue Of Products Sold In The Year 2015

    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.

     

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

     

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

     

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

     

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

     

    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.

    Figure 10 - Variation Of Revenue Of Products Sold In The Year 2016

    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.

     

    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.

     

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

     

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

     

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

     

    = 27761.003 million USD

     

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

    Figure 11 - Variation Of Revenue Of Products Sold In The Year 2017

    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.

     

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

     

    \(y=1.03^x\)

     

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

     

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

     

    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.

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  • Figure 12 - Variation Of Revenue Of Products Sold In The Year 2018

    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.

     

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

     

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

     

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

     

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

     

    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.

    Figure 13 - Variation Of Revenue Of Products Sold In The Year 2019

    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.

     

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

     

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

     

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

     

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

     

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

     

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

     

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

     

    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 cost17 of the products sold each year from the year 2010 to 2019.

    Figure 14 - Variation Of Cost Of Products Sold In The Year 2010

    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.

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  • Figure 15 - Variation Of Cost Of Products Sold In The Year 2011

    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.

    Figure 16 - Variation Of Cost Of Products Sold In The Year 2012

    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.

    Figure 17 - Variation Of Cost Of Products Sold In The Year 2013

    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.

     

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

     

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

     

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

     

    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.

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  • Figure 18 - Variation Of Cost Of Products Sold In The Year 2014

    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):

     

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

     

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

     

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

     

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

     

    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.

    Figure 19 - Variation Of Cost Of Products Sold In The Year 2015

    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):

     

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

     

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

     

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

     

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

     

    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.

    Figure 20 - Variation Of Cost Of Products Sold In The Year 2016

    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.

     

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

     

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

     

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

     

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

     

    = 12280.854 million USD

     

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

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  • Figure 21 - Variation Of Cost Of Products Sold In The Year 2017

    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):

     

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

     

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

     

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

     

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

     

    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.

    Figure 22 - Variation Of Cost Of Products Sold In The Year 2018

    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):

     

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

     

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

     

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

     

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

     

    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.

    Figure 23 - Variation Of Cost Of Products Sold In The Year 2019

    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):

     

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

     

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

     

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

     

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

     

    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.

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  • Determination of marginal cost and total cost

    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
    Figure 24 - Determination Of Marginal Profit And Total Profit From The Year 2010-2019

    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
    Figure 25 - Raw Data For Number Of Employees, Marginal Profit And Total Profit From The Year 2010-2019

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

    Determination of correlation and regression

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

    Conclusion

    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.
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  • Reflection

    Strength

    • 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.

    Weakness

    • 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.
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  • Further scope

    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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    Adib, PARVIZ, and DAVID Hurlbut. “Chapter 7 - Market Power and Market Monitoring.” Competitive Electricity Markets, edited by Fereidoon P. Sioshansi, Elsevier, 2008, pp. 267–96. ScienceDirect, https://doi.org/10.1016/B978- 008047172-3.50011-8.

     

    Arkes, Jeremy. Regression Analysis : A Practical Introduction.

     

    Athiyarath, Srihari, et al. “A Comparative Study and Analysis of Time Series Forecasting Techniques.” SN Computer Science, vol. 1, no. 3, May 2020, p. 175. DOI.org (Crossref), https://doi.org/10.1007/s42979-020-00180-5.

     

    James, Emily. What Is Correlation Analysis? Definition and Exploration.

     

    https://blog.flexmr.net/correlation-analysis-definition-exploration.%20Accessed%2011%20Sept.%202021.

     

    Khouja, Moutaz, and Stephanie S. Robbins. “Optimal Pricing and Quantity of Products with Two Offerings.” European Journal of Operational Research, vol. 163, no. 2, June 2005, pp. 530–44. DOI.org (Crossref), https://doi.org/10.1016/j.ejor.2003.10.036.

     

    “Marginal Revenue (MR) Definition.” Investopedia,

     

    https://www.investopedia.com/terms/m/marginal-revenue-mr.asp. Accessed 11 Sept. 2021.

     

    Mason, S. J., and G. M. Mimmack. “The Use of Bootstrap Confidence Intervals for the Correlation Coefficient in Climatology.” Theoretical and Applied Climatology, vol. 45, no. 4, 1992, pp. 229–33. DOI.org (Crossref), https://doi.org/10.1007/BF00865512.

     

    “Understanding Cost of Goods Sold – COGS.” Investopedia,

     

    https://www.investopedia.com/terms/c/cogs.asp. Accessed 18 Sept. 2021.

     

    “What Is the Relationship Between Marginal Revenue and Total Revenue?” Investopedia, https://www.investopedia.com/ask/answers/033115/what-relationship- between-marginal-revenue-and-total-revenue.asp. Accessed 11 Sept. 2021.

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