Take for example an entrepreneur starting up a music store that both produces and sells brass instruments. The owner, Joe, has designed a new professional grade euphonium and wants to mass produce them and sell them online for maximum profit. Joe has estimated that his costs are going to be: $900,000 for manufacturing set-up, promotion and advertising costs $3,000 to make each euphonium Joe has also undertaken extensive research into his market and based on his competitors, expects sales to follow a “Demand Curve” similar to: Unit Sales 350,000 – BOP; where P equals price.
If Joe sets the price at $0, he is giving way all the Euphoniums for free, or, by selling them at $4000, he will sell 30,000 euphoniums. However, the question still remains, what is the best price for Joe to sell his euphoniums and how much profit should he make? This is where the quadratic function and the parabola comes in. Joe must first come up with a quadratic expression. How much he sells depends on the price of the product.
Experimenting with the equation, we use price (P) as the variable: Unit Sales Sales in Dollars = Units x Price = Costs profit = sales – costs = (A quadratic equation) We now have a quadratic equation that can be solved by completing the square. (nearest whole number) The above equation shows that Joe will earn no profit when the price is $4366. 50 or $3008. 50. This means that maximum profit will be exactly Halfling. ‘ay between, shown on our parabola: Where the maximum point of the parabola is, we have maximum profit directly in line with the optimal sale price.
Therefore, as shown in figure 3, the ideal sales price is $3687. 50. This means: Unit Sales profit = Sales -? Costs = Through the use of a parabola and quadratic expression, Joe has been able to determine the optimum sales price for his product, $3687. 50, and estimate hat it will be quite a profitable business venture, earning a total profit of $36,912,500. The use of parabola for predicting sales can therefore be useful not only for new businesses or entrepreneurs to determine if a business venture will be profitable, but for existing businesses looking for renewal or growth in the development of new products.
Domain and Range Issues For any parabola, we can find the domain and range where: Domain: set of all possible x – values Range: set of all possible y -? values There are a number of restrictions on the domain: Cannot divide by zero Cannot take a negative root For the above parabola used in the above example, we can change the values of p to represent x, therefore, , and can find the domain and range where: Domain None of the above restrictions apply. Any real value of x can be used.