Problem 57 - Calculus and its applications

Question

A company determines that monthly sales \(S(t),\) in thousands of dollars, after \(t\) months of marketing a product are given by \(S(t)=2 t^{3}-40 t^{2}+220 t+160\) a) Find \(S^{\prime}(1), S^{\prime}(2),\) and \(S^{\prime}(4)\). b) Find \(S^{\prime \prime}(1), S^{\prime \prime}(2),\) and \(S^{\prime \prime}(4)\). c) Interpret the meaning of your answers to parts (a) and (b).

Step-by-Step Solution

Answer

The sales rate increases initially but then starts to decrease over time. The decrease is accelerating since the second derivative is negative.

Step 1: Calculate the first derivative S'(t)

To find the first derivative, differentiate the function \(S(t) = 2t^3 - 40t^2 + 220t + 160\) with respect to \(t\). The power rule states that the derivative of \(t^n\) is \(nt^{n-1}\).- The derivative of \(2t^3\) is \(6t^2\).- The derivative of \(-40t^2\) is \(-80t\).- The derivative of \(220t\) is \(220\).- The derivative of the constant \(160\) is \(0\).Combining these, we get:\[S'(t) = 6t^2 - 80t + 220\]

Step 2: Evaluate S'(t) at given points

Substitute \(t = 1\), \(t = 2\), and \(t = 4\) into \(S'(t) = 6t^2 - 80t + 220\) to find \(S'(1), S'(2),\) and \(S'(4)\).- For \(S'(1)\): \[S'(1) = 6(1)^2 - 80(1) + 220 = 6 - 80 + 220 = 146\]- For \(S'(2)\): \[S'(2) = 6(2)^2 - 80(2) + 220 = 24 - 160 + 220 = 84\]- For \(S'(4)\): \[S'(4) = 6(4)^2 - 80(4) + 220 = 96 - 320 + 220 = -4\]

Step 3: Calculate the second derivative S''(t)

To find the second derivative, differentiate \(S'(t) = 6t^2 - 80t + 220\) with respect to \(t\).- The derivative of \(6t^2\) is \(12t\).- The derivative of \(-80t\) is \(-80\).- The derivative of the constant \(220\) is \(0\).Thus, the second derivative is:\[S''(t) = 12t - 80\]

Step 4: Evaluate S''(t) at given points

Substitute \(t = 1\), \(t = 2\), and \(t = 4\) into \(S''(t) = 12t - 80\) to find \(S''(1), S''(2),\) and \(S''(4)\).- For \(S''(1)\): \[S''(1) = 12(1) - 80 = 12 - 80 = -68\]- For \(S''(2)\): \[S''(2) = 12(2) - 80 = 24 - 80 = -56\]- For \(S''(4)\): \[S''(4) = 12(4) - 80 = 48 - 80 = -32\]

Step 5: Interpretation of results

The first derivative \(S'(t)\) represents the rate of change of sales with respect to time. - At \(t = 1\), the sales rate is increasing by \(146\) thousand dollars per month.- At \(t = 2\), the rate of increase slows down to \(84\) thousand dollars per month.- At \(t = 4\), sales are decreasing as the rate is \(-4\) thousand dollars per month.The second derivative \(S''(t)\) shows the concavity of the sales curve or how the rate of change itself is changing.- Negative values at \(t = 1, 2,\) and \(4\) indicate the rate of increase is slowing or sales are declining faster.

Key Concepts

Rate of Change

In calculus, the rate of change is a fundamental concept that helps us understand how one quantity changes with respect to another. It shows us the speed of change, providing insights into how a system or a function behaves over time. Think of it as the speed of a car, which tells us how fast the car is moving at different times.

• The rate of change can have positive, negative, or zero values. A positive rate means the quantity is increasing, while a negative rate suggests it's decreasing. A zero rate of change implies no change is occurring.
• In sales analysis, the rate of change of sales can reveal if the sales are growing or shrinking over a specific period. For instance, a positive rate indicates increased sales, which is good for business, whereas a negative rate could mean something is going wrong.
• Calculating the rate of change involves finding the derivative of a function, which brings us to the concept of derivatives in calculus.

Derivatives

Derivatives play a crucial role in calculus. They provide valuable information about the behavior of a function by showing how the function's output changes with a small change in input. To find the derivative of a function like sales over time, we apply rules of differentiation.

• The power rule is one such differentiation method. It states that the derivative of any term like \(t^n\) is \(nt^{n-1}\).
• Derivatives help calculate the instantaneous rate of change, which is extremely useful for real-time insights. For instance, in our sales problem, the first derivative \(S'(t)\) gives us the rate at which sales are changing at any given time \(t\).
• Meanwhile, the second derivative \(S''(t)\) tells us about the rate of the rate of change. This can explain whether sales growth is accelerating or decelerating.

Derivatives, thus, are tools that allow us to predict trends, optimize functions, and understand the intricate dynamics of various phenomena like sales.

Sales Analysis

When examining sales over time, such as the given problem, sales analysis becomes an essential part of understanding business performance. By analyzing the mathematical function representing sales growth, we can gain insights into market dynamics and adjust strategies accordingly.

• Sales analysis using calculus allows businesses to foresee potential drops or rises in revenue. For example, a business might find that at \(t=4\) months, sales are on a downward path due to a calculated negative rate of change. This information is vital for adjusting marketing strategies or product offerings.
• The first derivative \(S'(t)\) informs how aggressive or conservative sales campaigns should be. It can show the precise moment when to push more in advertising or when to pull back.
• Furthermore, analyzing the second derivative \(S''(t)\) helps companies understand how to future-proof their strategies by knowing if the pace of change is sustainable or needs alteration.

In summary, calculus applications in sales analysis provide a data-driven approach to decision-making, helping companies maintain competitiveness and profitability.