Problem 68 - Calculus an Applied Approach

Question

Sales Analysis 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 derivative of the function \(M(t)\) is \(M^{\prime}(t) = 300(1-t^{2}) / (t^{2}+1)^{2}\). At 3 months, \(M(3) = 72\) memberships sold and \(M^{\prime}(3) = -60\) memberships per month. At 24 months, \(M(24) = 39.37\) memberships sold and \(M^{\prime}(24) = -0.31\) memberships per month.

Step 1: Find \(M^{\prime}(t)\)

The function \(M(t)\) is composed of a fraction and a constant. To find its derivative, we need to use the Quotient Rule for differentiation, which states that the derivative of a quotient \(u/v\) is \((vu'-uv')/v^{2}\). We let our \(u = 300t\) and our \(v = t^2 + 1\), resulting in the derivative \(M^{\prime}(t)\) as \((300(t^{2}+1) - 2t*300t) / (t^{2}+1)^{2}\) simplifying to \(M^{\prime}(t) = 300(1-t^{2}) / (t^{2}+1)^{2}\).

Step 2: Find \(M(3)\) and \(M^{\prime}(3)\) and interpret the results

Substitute \(t = 3\) into the function \(M(t)\) and its derivative \(M^{\prime}(t)\). We get \(M(3) = 300*3/(3^{2}+1) + 8 = 72\) and \(M^{\prime}(3) = 300(1-3^{2}) / (3^{2}+1)^{2} = -60\). These results mean that three months after the center opened, they sold 72 memberships, and the rate of change of sales at the third month was -60 memberships per month.

Step 3: Find \(M(24)\) and \(M^{\prime}(24)\) and interpret the results

Similarly, when we substitute \(t = 24\) into our function \(M(t)\) and its derivative \(M^{\prime}(t)\), we get \(M(24) = 300*24/(24^{2}+1) + 8 = 39.37\) and \(M^{\prime}(24) = 300(1-24^{2}) / (24^{2}+1)^{2} = -0.31\). These results mean that twenty-four months after the center opened, they sold approximately 39 new memberships, and the rate of sales was decreasing at 0.31 memberships per month.

Key Concepts

Quotient Rule

The Quotient Rule is a handy tool in calculus used to find the derivative of a function that is a ratio of two other functions. If we have a function expressed as a fraction \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( t \), the derivative is obtained through the Quotient Rule.

• The derivative is given by \( \left(\frac{vu' - uv'}{v^2}\right) \).
• Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

In our exercise, we are given the function \( M(t) = \frac{300t}{t^2+1} + 8 \). To apply the Quotient Rule, we identify:

• \( u = 300t \) and \( v = t^2 + 1 \), making \( u' = 300 \) and \( v' = 2t \).
• Using the rule, the derivative \( M'(t) \) simplifies to \( \frac{300(t^2 + 1) - 2t(300t)}{(t^2 + 1)^2} \), further reduced to \( \frac{300(1 - t^2)}{(t^2 + 1)^2} \).

This method allows us to differentiate the given function accurately, aiding in the analysis of sales behavior over time.

Derivative Interpretation

Understanding derivatives in a practical manner involves interpreting what the numerical results mean in real-world terms. The derivative of a function at a certain point provides the rate of change of the function's value with respect to its input variables.
In our exercise:

• \( M'(3) = -60 \) indicates that three months into operation, the sales of memberships are decreasing by 60 members per month.
• \( M'(24) = -0.31 \) shows that two years after opening, the decline in sales pace significantly, with a decrease of just 0.31 memberships per month.

These results help us understand trends such as initial high demand tapering off over time, possibly due to market saturation or competition. Thus, derivatives not only aid in mathematical computations but in understanding business dynamics as well.

Function Evaluation

Function evaluation is the process of substituting specific values into a function to obtain an output, helping us interpret the function's behavior at given inputs.
For the function \( M(t) \) in our exercise, evaluating:

• \( M(3) = 72 \) signifies that three months post-opening, the gym sold 72 memberships.
• \( M(24) \approx 39.37 \) indicates that after two years, sales dropped to approximately 39 memberships per month.

By calculating these, you grasp the changing demand for gym memberships over time. Initial high sales tapering to lower numbers, aligns with natural business cycles or changes in consumer interest.
Function evaluation bridges the gap between abstract mathematical modeling and tangible real-world insights, providing concrete data for decision-making.