Problem 61 - Algebra and Trigonometry with Analytic Geometry

Question

The demand for a commodity usually depends on its price. If other factors do not affect the demand, then the quantity \(Q\) purchased at price \(P\) (in cents) is given by \(Q=k P^{-c}\), where \(k\) and \(c\) are positive constants. If \(k=10^{5}\) and \(c=\frac{1}{2}\), find the price that will result in the purchase of 5000 items.

Step-by-Step Solution

Answer

The price \( P \) is 400 cents.

Step 1: Identify the problem variables

We are given the demand relationship for quantity \( Q \) and price \( P \), specifically \( Q=k P^{-c} \). The constants are \( k = 10^5 \) and \( c = \frac{1}{2} \), and we need \( Q = 5000 \). The task is to find the price \( P \).

Step 2: Substitute known values into the formula

Insert the given values into the equation: \( 5000 = 10^5 \cdot P^{-\frac{1}{2}} \). This gives us an equation linking \( P \) with \( 5000 \) items.

Step 3: Solve for \( P^{-\frac{1}{2}} \)

To isolate \( P^{-\frac{1}{2}} \), divide both sides by \( 10^5 \): \[ P^{-\frac{1}{2}} = \frac{5000}{10^5} = 0.05 \].

Step 4: Change the exponent to solve for \( P \)

We need \( P \), so we raise both sides to the power of \(-2\): \( (P^{-\frac{1}{2}})^{-2} = (0.05)^{-2} \).

Step 5: Simplify the expression

Calculate \( 0.05^{-2} \), which means \( \frac{1}{0.05^2} \). First, calculate \( 0.05^2 = 0.0025 \), then \( \frac{1}{0.0025} = 400 \). Thus \( P = 400 \).

Key Concepts

Demand Equation and Its Role

In economics, the demand equation is pivotal in modeling how a consumer behaves at different price points for a product. It helps you understand how much of a commodity is likely to be bought when the price is changed, assuming all other factors remain constant. For instance, consider the equation given: \[ Q = k P^{-c} \]Here,

• \(Q\) symbolizes the quantity demanded.
• \(k\) is a constant that provides scale; in our case, \(k = 10^5\).
• \(P\) represents the price of the commodity.
• \(c\) is a constant related to how demand changes with price, and here \(c = \frac{1}{2}\).

This demand equation describes an inverse relationship between quantity and price, which is common: as price increases, demand tends to fall. Conversely, as the price decreases, demand usually climbs. This information is important for businesses and economists alike, as it helps predict consumer behavior when setting prices.

The Art of Constant Solving

When dealing with equations in economics or mathematics, constant solving is essential. In the given problem, constants \(k\) and \(c\) help simplify the equation to find the unknown variable—in this case, price \(P\). Let's break down the constant solving process using the information:The equation given is: \[ 5000 = 10^5 \cdot P^{-\frac{1}{2}} \]To solve for \(P\), your task includes isolating the term \(P^{-\frac{1}{2}}\). Here's how you do this:

• Divide both sides of the equation by \(10^5\), simplifying the equation to \[ P^{-\frac{1}{2}} = \frac{5000}{10^5} \]
• Calculate the division to find \( P^{-\frac{1}{2}} = 0.05 \)

Such constant solving ensures clarity and provides the basis for moving forward with more complex math needed to solve for \(P\). It is this careful, step-by-step approach that makes the larger problem easier to unravel.

Exponent Manipulation in Equations

Exponent manipulation is a key mathematical tool, especially when faced with complex problems where powers are involved—in this case, to find the value of \(P\). The given problem hinges on manipulating the negative exponent to solve for \(P\).In the solution, we know:\[ P^{-\frac{1}{2}} = 0.05 \]Next, to solve for \(P\), you need to remove the negative exponent. Raising to the power of the reciprocal of this exponent can help do this:\[ (P^{-\frac{1}{2}})^{-2} = (0.05)^{-2} \]This operation essentially "flips" the exponent and transitions \(P\) into a form where it can be explicitly solved. Calculate this:

• (0.05)^{-2} calculated as \(\frac{1}{0.05^2} = \frac{1}{0.0025} = 400 \)

Thus, this manipulation reveals that \(P = 400\), as calculated earlier.Understanding exponent manipulation allows one to solve many similar mathematical challenges effectively, providing a firm foundation in mathematical problem-solving.