Question

Profit The demand and cost equations for a stethoscope are given by \(p=140-0.0001 x\) and \(C=80 x+150,000\) where \(p\) is the unit price (in dollars), \(C\) is the total cost (in dollars), and \(x\) is the number of units. The total profit \(P\) (in dollars) obtained by producing and selling \(x\) units is given by \(P=R-C=x p-C\) Try to determine a price \(p\) that would yield a profit of \(\$ 9\) million, and then use a graphing utility to explain why this is not possible.

Short answer

To yield a profit \(P\) of $9 million is not possible for the presented demand and cost equations. After solving the equations, it's found that the price \(p\) becomes negative, which is not feasible in this context. The negative pricing indicates that the cost is too high to produce and sell enough units to create $9 million profit given the current demand for the product (negative price is not viable in the economic sense). When plotting the equations using a graphing utility, it's visible that the unit cost and the demand don't allow for a profit of $9 million.

Step-by-step solution

Understand the problem

The profit equation given is \(P=R-C=x p-C\) where \(P\) is the profit, \(R\) is the total revenue, \(C\) is the total cost, \(p\) is the unit cost and \(x\) is the number of units. Here, it is asked to find the price \(p\) to yield a profit \(P\) of $9 million.

Define the Revenue

According to the problem, Revenue which is \(R\) is derived from the multiplication of the number of units \(x\) and the unit price \(p\). So, \(R = x*p = x*(140-0.0001x)\).

Substitute the equations into the profit formula

Now, substitute the Revenue \(R\) and Cost \(C\) equations into the profit formula, we get \(P=(140-0.0001x)*x-(80x +150,000)\). Since we are told that the profit \(P\) should be $9 million, that gives us \(9,000,000=(140-0.0001x)*x-(80x +150,000)\).

Solve for \(x\)

Solving the above equation gives us a quadratic equation that needs to be solved by setting it equal to zero. Solving the quadratic equation for \(x\) yields the number of units that need to be produced and sold in order to reach a $9 million profit.

Find the price \(p\)

After solving the quadratic equation, we obtain two values for \(x\). Substituting each \(x\) value into the pricing equation \(p=140-0.0001x\), we can find two corresponding prices \(p\) for each \(x\). However, because \(p\) can't take a negative value, discard the \(x\) value that leads to a negative price.

Validate the Price using a graphing utility

After finding the valid unit price which gives the profit of $9 million, use a graphing utility to illustrate the relations between unit price \(p\), cost \(C\), revenue \(R\), number of units \(x\) and profit \(P\). By plotting these relations, one can see why it's impossible to achieve a profit of $9 million with the given equations.

Key concepts

Quadratic Equations

Understanding how to solve quadratic equations is pivotal to many areas of algebra, including business applications like profit maximization. A quadratic equation is a second-degree polynomial typically written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(^x\) is the unknown variable. These equations often arise when trying to determine optimal values, such as pricing strategies that maximize profit.

To solve such equations, one may use factoring, completing the square, the quadratic formula, or graphing. In the context of the profit maximization problem provided, we rearranged the profit equation into a quadratic format, which allowed us to solve for the number of units, labeled here as \(x\), necessary to achieve a specific profit goal. Students may benefit from practicing the quadratic formula, as well as visualizing the solutions on a graph, to better grasp the concept of maximum and minimum points, which correspond to the peak profit points in business scenarios.

Cost-Revenue-Profit Analysis

The concepts of cost, revenue, and profit are fundamental to understanding the financial health of a business. In algebraic terms, the relationship between these three can be defined as \(Profit = Revenue - Cost\). The cost function \(C\) typically represents the total expenses incurred in the production and delivery of goods or services, while the revenue function \(R\) reflects the total income from sales.

The goal of cost-revenue-profit analysis is to find the optimal balance between pricing and production that maximizes profit. By developing and solving a system of equations that represent these relationships, business owners and managers can make informed decisions. In the exercise we examined, forming and solving an algebraic equation allowed us to explore the feasibility of reaching a particular profit target, which, as proven mathematically and graphically, may not always be possible due to the constraints imposed by the cost and demand functions.

Graphing Utility Applications

Graphing utilities, such as calculators and software, are invaluable tools that offer visual insights into complex algebraic equations. They enable students to see the graphical representation of equations like those in the profit maximization problem. Particularly, they are used to illustrate how the unit price affects both revenue and cost, and in turn, profit.

By plotting the demand, cost, and profit functions, students can identify key points such as the break-even point, where the revenue and cost curves intersect, and the profit maximization point, where the profit curve reaches its peak. Furthermore, graphing utilities can demonstrate why certain objectives, like the $9 million profit in the given scenario, may be unattainable under existing conditions, thus providing a more intuitive understanding of the solutions derived algebraically.