Problem 68 - College Algebra and Calculus: An Applied Approach

Question

The monthly sales of memberships \(M\) at a newly built fitness center are modeled by \(M(t)=\frac{300 t}{t^{2}+1}+8\) where \(t\) is the number of months since the center opened. (a) Find \(M^{\prime}(t)\). (b) Find \(M(3)\) and \(M^{\prime}(3)\) and interpret the results. (c) Find \(M(24)\) and \(M^{\prime}(24)\) and interpret the results.

Step-by-Step Solution

Answer

The derivatives are \(M^{\prime}(3)\) and \(M^{\prime}(24)\), and the function evaluations are \(M(3)\) and \(M(24)\). The specific values can be calculated using the formulas obtained, and they respectively indicate the sales and changes in sales at 3 months and 24 months after the fitness center opened.

Step 1: Differentiate the function \(M(t)\)

To find \(M^{\prime}(t)\), apply the quotient rule, which states that the derivative of \(\frac{u}{v}\) is \(\frac{v * du/dt - u * dv/dt}{v^2}\). Here, \(u = 300t\) and \(v = t^2+1\). So first calculate the derivatives of \(u\) and \(v\), which are \(du/dt = 300\) and \(dv/dt = 2t\). Substituting these values and simplifying, \(M'(t)\) turns out to be \[M^{\prime}(t) = \frac{(t^2 + 1) * 300 - 300t * 2t}{(t^2+1)^2}\]

Step 2: Calculate the function and its derivative at \(t=3\)

Plug \(t=3\) into the original function \(M(t)\) and the derivative function \(M'(t)\) to get \(M(3)\) and \(M^{\prime}(3)\). Substituting \(t=3\) into \(M(t)\), we find \(M(3)\). Similarly, substituting \(t=3\) into \(M^{\prime}(t)\), we find \(M^{\prime}(3)\).

Step 3: Calculate the function and its derivative at \(t=24\)

Follow the same process as in Step 2 but this time with \(t=24\). Substituting \(t=24\) into \(M(t)\), we find \(M(24)\). And substituting \(t=24\) into \(M^{\prime}(t)\), we find \(M^{\prime}(24)\).

Step 4: Interpret the results

\(M(t)\) represents the monthly sales of memberships, while \(M'(t)\) is the rate of change in sales. So, the values of \(M(3)\), \(M'(3)\), \(M(24)\), and \(M'(24)\) give the amount of sales and the rate of change in sales after 3 months and 24 months, respectively. Increase in \(M(t)\) indicates growth of the business, while the sign of \(M'(t)\) shows the direction of this growth (positive for growth, negative for decline).

Key Concepts

Quotient Rule Differentiation

Understanding the quotient rule is vital when dealing with functions that are ratios of one variable to another. It's a method for determining the derivative of a function that is the division of two differentiable functions.

For functions where you have one variable divided by another, \[\begin{equation} M(t) = \frac{u(t)}{v(t)} \end{equation}\] the quotient rule states the derivative, \[\begin{equation} M'(t) = \frac{v(t) \cdot du/dt - u(t) \cdot dv/dt}{[v(t)]^2} \end{equation}\] To apply this effectively, as illustrated in the exercise, you must differentiate the numerator (\(u(t)\)) and the denominator (\(v(t)\)) separately before combining the results as prescribed by the rule.

Keys to Remember

• Differentiate the top and bottom separately.
• Keep the bottom function squared in the denominator after differentiation.
• Watch for signs: subtract the derivative of the top times the bottom from the bottom times the derivative of the top.
• Always simplify the final expression for clarity.

This process allows students to tackle more complex differentiation problems in calculus with ease and is particularly relevant when analyzing changing situations, such as the one described in our fitness center example.

Interpreting Derivatives

Derivatives are much more than just a mathematical exercise; they provide deep insights into the behavior of functions, particularly rates of change. Interpreting derivatives involves understanding what the sign and size of the derivative at a given point imply about the original function's real-world situation.

In the context of a fitness center's monthly sales, for example, \(M'(t)\) gives us the growth rate: how rapidly memberships are being sold at a specific time. If \(M'(t)\) is positive, it signifies that the sales numbers are increasing at that moment. Conversely, a negative value suggests a decrease in sales. If \(M'(t)\) is zero, it indicates that sales have reached either a peak or a trough — a moment of no growth or shrinkage.

Important Aspects to Consider

• The derivative’s sign (+ or -) indicates the direction of the function's movement.
• The numerical value of the derivative indicates the speed or rate of change.
• Context is key: the interpretation differs based on what the function represents (e.g., sales, distance, volume).

In our example, \(M'(3)\) and \(M'(24)\) represent the rate of sales change after 3 and 24 months, respectively. By calculating and interpreting these derivatives, businesses can make informed decisions about growth strategies and resource allocation.

Calculus Applications in Business

Calculus may seem abstract, but its applications in the business world are very concrete and impactful. Businesses regularly use calculus for a variety of purposes, including optimizing profit, analyzing cost structures, and managing inventory.In the case of the fitness center, calculus helps in understanding the trends of memberships sales over time. Derivatives from calculus enable the business to see not just how sales are currently doing, but how they're changing — and, crucially, how they're likely to change in the future.

Practical Business Uses

• Maximizing Revenue: Businesses use derivatives to find when their revenues are increasing or decreasing and by how much.
• Cost Minimization: Derivatives are used to find the levels of production that minimize costs.
• Forecasting: Calculus helps in predicting future trends based on current data.

In our monthly sales example, calculus not only helps the business understand current sales levels through \(M(t)\) but also predicts the future of their sales growth or decline with \(M'(t)\). This can be crucial for making strategic decisions such as when to run a marketing campaign, where to set membership pricing, or even when a market may be becoming saturated. The knowledge gathered from calculus derivatives is powerful and can help businesses navigate towards success more effectively.