How to maximize profit when considering other costs.

How can derivatives be used to determine the quantity of production that maximizes a company’s total profit?When learning about derivative functions, I became curious about the applications of them. Through researching about the application of derivatives functions, I decided to base this Mathematical Exploration on optimization in business, finding an alternative with the most cost effective or highest achievable performance under the given constraints, by maximizing desired factors and minimising undesired ones.I decided to explore the different situations where derivative functions could be applied the first being the situation of a storage space rental company and whether or not they should rent all of their spaces to maximize profit when considering other costs. Extending the model, I decided to explore the production costs of a wrench company and how to minimize them and also finding the rate of change of the cost at x= a number. b Derivatives in the business world is useful in that we can easily compute how much to sell or produce when considering other expenses such as labor cost, marginal cost, and production cost. This exploration aims to develop a model for optimization for Nike in how many tennis shoes they should sell to receive the most profit.To accurately model optimization of Nike in how many tennis shoes they should sell to maximize their profit and minimize their costs, I decided to examine simple scenarios of optimization equations to be able to understand derivative applications for myself. I also decided to determine the factors related to optimization scenarios which are salient to the model: one being as mentioned above the production costs, marginal costs, and maintenance costs. When looking at the model, it is important to note the constraints of the equation. Selling or producing the most goods does not always mean most profit when considering other costs. First ExampleTo accurately depict the first example consider the costs mentioned above like production costs, marginal costs, and maintenance costs. A storage space rental company had 300 spaces to rent. If they rent x spaces their monthly profit in dollars is given by P(x)= -6×2+ 3600x-70,000. To find out the number of apartments spaces to rent first take the derivative of the points that fall within the constraint 0?x?300For the first variable multiply the exponent and the constant and subtract 1 from the exponent and get -12x the derivative of the second term is 2600 and the derivative of a constant by itself with no variable is 0.So the derivative of product equation can be written as -12x+3600 or 3600-12x.The derivative equation can be solved using basic rules 360012=300 In this scenario, the company should rent 300 of its storage spaces. The maximum of the constraint was the best option because the maintenance cost was relatively low. Second ExampleA wrench company is able to produce 80,000 wrenches per day and the total daily cost of producing X number of wrenches in a day is given by, 270,000 + .1x + 22000000x. To minimize the production costs we must set the constraint 0?x?80,000. Next, find the derivative of the equation: .6 + 2’160’000’000x2then solve for x to get ±60,000. The negative derivative value does not make sense in this scenario because it does not fall within the constraints but the positive derivative solution does fall within the constraint therefore the wrench company should produce 60,000 wrenches to minimize the production costs. Third ExampleThe weekly production costs of the wrench company is model by C(x)= 400+ 250x-0.08x2To figure the rate of change of the cost at x=200, take the derivative of the function to get 250-0.16x. Next, insert 200 into the equation for x to get 218. The value of the rate of change of the cost at x= 200 is 218. “The Marginal cost is defined as the cost to the produce an additional item. The Marginal cost is approximated by the rate of change of the cost function.” That is why finding the rate of change is important. Modeling Nike’s total profit for producing x number of unitsThe profit function is given by: P(x)=R(x)-C(x)=xp(x)-C(x). The profit function represents the total revenue, R(x), from producing x number of units subtracted by the total cost C(x), the cost of producing x number of units, and P(x) the profit or loss by selling and producing x number of units. The following table shows the average production costs for nike to make a tennis shoe that sells in retail for 100 dollars..Per ShoeCost Income StatementPercent of Revenue$25.00Factory FOB Cost $1.00Sea Freight and Insurance $2.50Duty $28.50Landed Cost (57% of Revenue) Cost of Goods57% $15.00 SG%$2.00 Tax4%$4.50 Profit9%$21.50Mark-up (43% of Revenue) 43% $50.00Wholesale Price Revenue 100%$50.00Retail Mark-up (100% of Revenue) $100.00Suggested Retail Based on this information, nike spends on average $28.50 to make one pair of sneakers. For each shoe sold the total revenue is $50.00. On average nike sells about 25 pairs of sneakers per second. To model how much profit nike will earn subtract the total cost from the total revenue. $50.00-$28.50=21.5 Nikes total profit per each pair of shoe sold is $21.502.) To find the total cost to produce 25 sneakers subsite 25 for x in the cost function C(x)=25(25)+1.00(25)+2.50(25)=$712.50 The total cost to produce 25 sneakers is $712.503.) For every one pair of shoes nike makes 50 dollars. To find out how much nike makes per 25 pairs substitute 25 in for 50 in the revenue equation.R(x)=50(25)=$1250For 25 pairs of shoes nike makes a total revenue $1,2504.) The profit function is calculated by subtracting the total cost from the revenueP(x)=R(x)-C(x)         $1250-$712.50=$537.50In 1 second Nike on average makes $537.50 per second5.) To find out the approximate profit after selling 25 pairs of sneakers subtract the Cost function from the revenue functionP(x)= 50x-(25x+1.00x+2.50x)21.5x= 21.5(26)= $559The approximate profit after the next pair of shoes after 25 pairs of shoes is $559. In one second the nike makes a financial gain of $559.To model how much nike apply the optimization concept to Nike’s business, first one must analyze and profit and revenue report to answer how much nike needs to sell to maximize their profit for a full year. From the values given one must create a total  revenue equation  and use a graphing calculator.  Creating  a cost function is also essential to understanding the optimization concept as it applies to Nike. Based on the total revenue function and the cost function, one must find a suitable profit function. To find the quantity that Nike needs to sell to create find the derivative of the profit function and confirm that the value produces a maximum using the second derivative test.   So as discussed above Nike makes a profit $21.50 for every pair of shoe sold. The total revenue equation reads TR= PxQ, p for price of selling a good, and q representing the quantity. For every pair of shoes nike sells them for $100. For one pair of shoes Nike makes a total revenue of 100. Substituting the price function 100q in for P, the total revenue equation will yield a quadratic equation 100q2Nikes reported first quarter revenue for the fiscal 2018 year was 9.1 billion and Nikes total expenses including the demand creation expense and the operating overhead expense totaled to be 2,856,000 dollars. .Profit is the financial gain of a business of from producing a good or selling a service, which is calculated by the difference between total revenue and total costs. Businesses gain a profit when the total revenue is greater than the total cost. To figure out how Nike can maximize its profit create a function that subtracts the expenses from the revenue and apply the first and second derivative test to ensure that the value is a maximum.100q-28.50qf'(q)= 200q-28.50200q-28.50=0q=14.25Insert values less than and greater than 14.25. After inserting the values greater and less than 14.25, the value does lead to a maximum. However, 14.25 is not a whole number, so Nike needs to produce 14 pairs of shoes to maximize profit. Bibliography”Mathematics & Statistic Tutor Perth – SPSS Help   .” Profit, Cost and Revenue,, Brendan. Sole Collector, Sole Collector, 20 Oct. 2016,’s How Many Pairs of Sneakers Nike Sells Every Second of the Day.” HYPEBEAST,”Applications of Derivatives to Business and Economics.” Math Hawaii,