Problem 82 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

DEMAND FOR PCs The quantity demanded per month, \(x\), of a certain make of personal computer (PC) is related to the average unit price, \(p\) (in dollars), of PCs by the equation $$ x=f(p)=\frac{100}{9} \sqrt{810,000-p^{2}} $$ It is estimated that \(t\) mo from now, the average price of a PC will be given by $$ p(t)=\frac{400}{1+\frac{1}{8} \sqrt{t}}+200 \quad(0 \leq t \leq 60) $$ dollars. Find the rate at which the quantity demanded per month of the PCs will be changing 16 mo from now.

Step-by-Step Solution

Answer

The rate at which the quantity demanded per month of the PCs will be changing 16 months from now is: \[dx/dt = \frac{-100*400}{9\sqrt{810000 - 400^2}} \cdot \frac{-200}{8(1+\frac{1}{8}\sqrt{16})^{2}\sqrt{16}} \approx 0.625 \text{ units/month}\]

Step 1: Understand the relationship

In the problem, the quantity demanded per month \(x\) is given by a function of price \(f(p)\), and the price is a function of time \(p(t)\). So, to find how the quantity demanded changes over time, the derivative of \(x\) with respect to time \(t\) must be found, not the price. Therefore, chain rule is needed, which is a rule for computing the derivative of the composition of two functions.

Step 2: Differentiate the given functions with respect to their variables

Firstly, find the derivative of the function \(f(p)\) with respect to \(p\). Since \[f(p) = \frac{100}{9}\sqrt{810000 - p^2}\] the derivative with respect to \(p\) can be found using the derivative rule for the square root function and the chain rule: \[f'(p) = \frac{-100p}{9\sqrt{810000 - p^2}}\] Similarly, find the derivative of \(p(t)\) with respect to \(t\): \[p'(t) = \frac{-200}{8(1+\frac{1}{8}\sqrt{t})^{2}\sqrt{t}}\]

Step 3: Apply the Chain Rule

Using the chain rule, which states that \((f(g(x)))' = f'(g(x)) \cdot g'(x)\), find the derivative of the demand with respect to time, which is given by \[f'(p(t)) \cdot p'(t)\] By substituting the calculated derivatives from the previous step, \[dx/dt = \frac{-100p(t)}{9\sqrt{810000 - p^2(t)}} \cdot \frac{-200}{8(1+\frac{1}{8}\sqrt{t})^{2}\sqrt{t}}\]

Step 4: Calculate the value

Finally, to find the rate of change of the quantity demanded at 16 months, substitute the given value of \(t = 16\) into the equation obtained in the previous step: By substituting \(t = 16\) into \(p(t) = \frac{400}{1+\frac{1}{8}\sqrt{16}}+200\), we get \(p(16) = 400\). Then substitute into the derivative, \(dx/dt = \frac{-100*400}{9\sqrt{810000 - 400^2}} \cdot \frac{-200}{8(1+\frac{1}{8}\sqrt{16})^{2}\sqrt{16}}\), and compute the result.

Key Concepts

Demand Function

In economics, a demand function is an essential tool used to describe how the quantity demanded of a good or service varies based on its price. Specifically, it tells us the relationship between price and quantity demanded.
For the problem at hand, the demand function is defined by the equation:

• \(x = f(p) = \frac{100}{9} \sqrt{810,000 - p^{2}}\)

This equation indicates how many PCs consumers want to buy each month based on the price \(p\).
In general, as the price of an item decreases, the quantity demanded increases, and vice versa. The provided function illustrates this principle, showing a mathematical representation of consumer behavior in the market for PCs.

Chain Rule

The chain rule is a fundamental technique in calculus that is used to find the derivative of a composition of two or more functions.
It is particularly useful when dealing with problems where one variable depends on another through an intermediate variable.
In practical terms, imagine you have two functions: \(y = f(u)\) and \(u = g(x)\). When you want to find \(\frac{dy}{dx}\), you use the chain rule, which states:

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)

In our exercise, the quantity demanded \(x\) is a function of price \(p\), which, in turn, is a function of time \(t\). To find how \(x\) changes over time, the chain rule helps us by combining these dependencies into a single derivative with respect to \(t\). This gives us a step-by-step tool to derive complex relationships.

Derivative Computation

Computing derivatives is a core part of calculus, enabling us to understand rates of change.
In this exercise, we deal with derivatives in two parts:

• The derivative of the demand function \(f(p)\) with respect to \(p\).
• The derivative of the price function \(p(t)\) with respect to \(t\).

Firstly, for \(f(p)\):

• Given \(f(p) = \frac{100}{9}\sqrt{810000 - p^2}\), the derivative \(f'(p)\) is \(\frac{-100p}{9\sqrt{810000 - p^2}}\).

Secondly, for \(p(t)\):

• Given \(p(t) = \frac{400}{1+\frac{1}{8}\sqrt{t}}+200\), the derivative \(p'(t)\) is \(\frac{-200}{8(1+\frac{1}{8}\sqrt{t})^{2}\sqrt{t}}\).

These derivatives provide the rate of change of each function with respect to their respective variables. By chaining them together, we can compute \(\frac{dx}{dt}\), the rate of change of quantity demanded over time.

Quantity Demanded

The term 'quantity demanded' refers to the amount of a product that consumers are willing and able to purchase at a given price, during a specific period.
It fluctuates as the price changes, illustrating the law of demand—a core principle in economics.
In our case, we're interested in how this quantity changes over time as prices evolve. This is expressed mathematically by taking the derivative of the demand function with respect to time.
To refine this into something actionable, consider:

• Understanding current consumer preferences.
• Analyzing external factors such as technological advances or seasonal trends.

These elements can cause shifts or movements along the demand curve, reflected in changing quantity demanded over time. Such analysis is vital for businesses to anticipate changes in market demand and adjust strategies accordingly.

Price Function

The price function \(p(t)\) is a mathematical representation predicting how price changes over time.
It's often a necessary feature in applied mathematics to model real-world scenarios, such as future pricing based on current trends or anticipated events.
In this exercise, the price function is:

• \(p(t) = \frac{400}{1+\frac{1}{8}\sqrt{t}}+200\)

This function anticipates the price of the PC at any given month \(t\), shaping our understanding of how demand will shift.
Such functions are indispensable for planning and forecasting. They can inform decisions about inventory, marketing, and pricing strategies. By analyzing \(p(t)\), we can utilize the derivative to predict the impact of price changes on demand, providing insights into consumer behavior and helping to optimize economic outcomes.