MATHEMATICS EXPLORATIONNFOR YANNDP 2ATOPIC: USING CALCULUS (Optimisation) TO DETERMINE THE MAXIMUM PROFIT OF A BUSINESS SUPERVISOR: MR.NGONGA DIVINETable of Content Introduction to Calculus and Optimisation.……………………………………………3 Rationale for topic…………………………………………..………………………….3 Methodology…………………………………………………………………………..4 Research Problem……………………………………………………………………..5 Solution…………………………………………………………………………………5 Revenue equations from graph…………………………………………………6 Maximising profit using equation 1……………………………………………7 Maximising profit using equation 2………..…………………………………..9 Conclusion……………………..……………………………………………………11 Comparing both results Evaluation References………………………………………….…………………….…………..12Introduction Optimisation is a mathematical method used in finding the optimal or relative maximum of a quantity using a function. This is done through the second derivative of the function under differential calculus i.e. f”(x). We know that at the turning point of the graphs of polynomial functions of degree 2, the first derivative of a function or f'(x) = 0, so to know where y is at its maximum or minimum we use the second derivative. For my mathematics exploration, optimisation instantly gave me an area to investigate-business. As a business student, I taught why not use optimisation to maximise something like revenue or sales in a business. Considering the fact that actual businesses do not reveal certain costs figures, for this exploration I will have to come up with figures for a business and proceed accordingly. In optimisation, we determine optimal value of an equation under a constraint or a limiting factor. The constraint in this case, will be the amount of cash I invest for a production of a good. The constraint figure will then be represented using the optimum equation: a+ b = (constraint). The value of the variables x and y will be found when proceeding with the optimisation procedure, and the value found will be the maximum amount that when you the two constants together, you will get the optimum amount of what you are trying to find. In this case, I am trying to find the maximum profit a business can make. Its formula goes thus: profit=revenue-costs. Inserting this equation for profit in the equation for maximisation will give a – b=c (constraint), where x represents revenue and y represents the costs. Using this equation I will be able to maximise the amount of profit that Nike will be able to make. Rationale Optimisation deals with maximisation of a certain quantity. During our calculus class, our teacher told us optimisation and its applications. Being a business student, it immediately came to my mind to maximise profit. I inquired myself on a way in which I could use optimisation methods to maximise profit of a business. That is when revenue and costs aspects came to my mind. How could I use revenue and costs to maximise the profit gained from a product of a business using optimisation? Also, because I was currently searching for a topic for my Mathematics exploration I immediately started working on it and so now let me proceed with methodologies involved that I will use in order to maximise the profit of Nike Inc.. MethodologyThe methodology here includes all the rules and procedures involved with optimisation. We can’t optimise the profit of a business if we don’t even know the procedures and rules we have to follow in order to solve the problem. The first thing to know is that optimisation involves using first and second derivatives of functions. The functions in this case will be the function for revenue i.e. r(x) and the function for cost i.e. c(x). These functions will be developed once the value for the constraint is found in the optimum equation given earlier i.e. a -b =c (constraint). c is substituted with the value of the constraint. Once this is done we derive the equations and functions for the revenue and costs which will be substituted in the formula for calculating profit given Profit = revenue – costs. We then proceed by subtracting costs from revenue to get a function for profit i.e. P(x). After the substitution, we then have to find the derivative of the function P(x) i.e. P'(x). This is because we know that where f'(x)=0, there is the existence of a relative minimum or maximum value that the function can have. So as we find the first derivative of the P(x) function, we equate it to 0.As we equate the first derivative of the profit function to 0, we get a value(s) for x i.e. k. This will be the value at which there is a relative maximum or minimum. In order to check for a relative maximum, we have to find the second derivative at each 0 of the first derivative i.e. We substitute k in the second derivative of the profit function P(x). We will be able to determine whether there is a relative maximum or minimum by seeing the final value we get from substituting k. If f”(x) < 0, then there is a relative maximum and if f"(x) > 0, then there is a relative minimum. Research problemMy exploration will be limited to only real revenue figures for Nike Inc. as they do not make other figures about costs and invested money available to the general public. That is why I shall come up with an amount for costs and invested money in order to proceed with my exploration. Nike from 2009-2017 invested an average of 2500 million US$ in the production and selling of footwear. Considering the fact that Nike doesn’t dispose its costs figures, I came up with a fix costs figure for a every unit sold (x). So let’s assume that the costs for selling footwear (x) in North America from 2009-2017 was constantly 1000 US$ (in millions of dollars). As for revenue, I will use the graph of Nike’s North American revenue for footwear from 2009-2017 in order to have an equation for revenue. . Solution Before anything else, I need to come up with an optimum equation. Nike invested an average of 2500 million US$ in the production and selling of footwear in North American, this means that it was limited by this amount while producing. So my constraint in this case will be 2500 million US$. The optimum equation will therefore be: a – b = 2500 (in millions) The next step is then to bring out the equations or function for the revenue and costs. Nike doesn’t make available its costs figures, so I will use the assumption made above. The fixed amount of 1000US$ (in millions) were paid for the production and selling of ever unit x of footwear. This will give 1000x For revenue, the information from the graph below shall give derive me an equation that I will use in the rest of the exploration. The information from the graph shows to us the revenue figures for Nike’s footwear in North America from 2009-2017. Using Microsoft excel I was able to bring out two equations; 1 linear and 1 exponential equations. They are as thus: x = 706, 18x – 1E + 6 (linear) and x = 5E – 88e. (Exponential)Because we have 2 equations, I will use the each of them and compare the final results to see which maximises profit the most.When we simplify the linear equation, we get: x = 706, 18x – 1 x 106. x= 706, 18x – 10000000 With this, we have the equation for revenue so the next step is to substitute it with the cost value in the profit formula. However, let’s assume that because of some accounting problems, Nike actually needed to pay the square of what it thought as cost per unit sold from 2009-2017 i.e. (1000)x2 This changes the costs value thus: (1000x)2 10002×2 1000000x2Moving to the profit formula: P = r – c Since we are working in terms of x we adapt the formula in terms of x thus: P(x) = r(x) – c(x) Substituting revenue and costs equation and function: P(x) = 18x – 10000000 – (1000000×2) P(x) = 18x – 10000000 – 1000000×2 P(x) = -1000000×2 + 18x – 10000000 Now after having found and simplified the function for profit in terms of x I will proceed by finding its first derivative. P'(x) = -2000000x + 18 – 0 P'(x) = -2000000 + 18In order to get the relative maximum profit, we then equate the profit function to 0. -2000000x + 18 = 0 -2000000x = -18 + 0 x= -18/-2000000 x= 0.009We have a value for x, we don’t know whether it is a relative maximum or minimum. Therefore, to solve this problem we substitute this value in the second derivative of the profit function. The end product that results after simplifying will determine whether it is a relative maximum or not as P”(x) < 0, indicates where there is a relative maximum and P"(x) >0, indicates where there is a relative minimum. P”(x) = 0 P”(x) = 0 . 0.009 P”(x) = 0P”(x) <0 indicates that there is a relative maximum value of x, but in this case P"(x)=0This means that P"(x) isn't less than zero (<0) nor is more than zero (>0). Rather it is equal to zero (= 0), implying that there is no relative maximum or minimum. The implications of this shall be explored more keenly in my conclusion. Now that I have used the first revenue equation from the revenue graph, I need to use the second equation that I derived which is an exponential equation. The exponential equation goes thus: x = 5E – 88e.After a bit of simplification and an manipulation it gives: x = 5 x 10-88 . e0.1042xWith this, we have the equation for revenue so the next step is to substitute it with the cost value in the profit formulaSince we are working in terms of x we adapt the formula in terms of x thus: P(x) = r(x) – c(x) Substituting revenue and costs equation and function: P(x) = 5 x 10-88 . e0.1042x – (1000000×2 ) Now after having found and simplified the function for profit in terms of x I will proceed by finding its first derivative. Looking at my function i.e. P(x) = 5 x 10-88 . e0.1042x – (1000000×2 ), I see that I wont be able to proceed with the first derivative like I did earlier. I realise that I will be able to differentiate 5 x 10-88 and 1000000×2 normally. However, for e0.1042x I will need to follow a different rule in finding out its first derivative.The rule goes thus: If y = ef(x) Then y’ = f'(x) . ef(x) Therefore, P'(x) = 0 . e0.1042x – (2000000x) Taking, e(0.1042x) Let (0.1042x) = f(x) If we follow the rule formula: y= e0.1042x y’= 0.1042 . e0.1042x Now when we substitute this back in the second derivative function of profit P”(x), we get: P'(x) = 0 x 0.1042 . e0.1042x – 2000000x P'(x) = 0 . e0.1042x – 2000000x P'(x) = 0 – 2000000x P'(x) = -2000000x In order to get the relative maximum profit, we then equate the profit function to 0. -2000000x = 0 -2000000x / -2000000 = 0/-2000000 x= 0We don’t know whether it is a relative maximum or minimum. Therefore, to solve this problem we substitute this value in the second derivative of the profit function. The end product that results after simplifying will determine whether it is a relative maximum or not as P”(x) < 0, indicates where there is a relative maximum and P"(x) >0, indicates where there is a relative minimum. P”(x) = -2000000(0) P”(x)= 0Looking at the outcome we see that the second derivative is equivalent to zero. This means for this second revenue equation, there is no relative maximum or minimum. This can be proven by checking for a relative maximum or minimum P”(0) = 0 (P”(x) is not less than zero i.e. < 0 to give a relative maximum) P"(0) = 0 (P(x) is not more than zero i.e. >0 to give a relative minimum)This means at this value there is a relative maximum value for x.Conclusion After having used the two equations for Nike’s footwear revenue from 2009-2017 together with anticipated costs and invested money figures I was able to deduct a potential method for maximising profit all of this under the rules of optimisation. However, since I used two equations, I have two results. Using the linear equation for revenue, P”(x) = 0, this meant that with this equation and the other figures, Nike could not maximise profit nor minimise it because when P”(x)< 0, then there is a relative maximum and when P"(x) > 0, then there is a relative minimum, however it was equal to zero i.e. P”(x) = 0.In the same light looking at the results from the second equation we see that there was also no method that Nike could to maximise its profit numbers based on the cost and revenue figures they had. This implies that Nike needs to change its and costs and revenue figures in a way that would favour the maximisation of profit and the minimisation of loss for the business.References G. Stephanie., (April 9, 2012), Optimisation problems in Calculus, www.statisticshowto.com/how-to-solve-optimisation-problems-in-calculus/ , retrieved November 10 2017 D. Paul (2017), Calculus I – Notes: Optimisation, www.tutorial.math.lamar.edu/Classes/Calcl/Optimisation.aspx, retrieved November 16 2017 B. Laurie, F. Jim, K. Ed, L.R. Paul, S. Jill, (2014), Mathematics Standard Level Course Companion: Limits and Derivatives; More on Extrema and optimisation problems, pages 240-249. Oxford University Press