Problem 56 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The quantity demanded each month of the Sicard wristwatch is related to the unit price by the equation $$ p=\frac{50}{0.01 x^{2}+1} \quad(0 \leq x \leq 20) $$ where \(p\) is measured in dollars and \(x\) is measured in units of a thousand. To yield a maximum revenue, how many watches must be sold?

Step-by-Step Solution

Answer

To yield maximum revenue, 10,000 watches must be sold.

Step 1: Find the Revenue Function

The revenue (R) is obtained by multiplying the price (p) by the quantity demanded (x). From the given equation, \(p = \frac{50}{0.01x^2 + 1}\). Now, let's find the revenue function (R): \(R(x) = p \cdot x = \frac{50}{0.01x^2 + 1} \cdot x\)

Step 2: Find the critical points of the Revenue Function

To find the maximum revenue, we need to determine the critical points of R(x) by taking its derivative with respect to x and setting it equal to 0. First derivative of R(x) with respect to x: \( R'(x) = \frac{d}{dx}\left(\frac{50x}{0.01x^2 + 1}\right)\) To find the derivative, we'll use the quotient rule. The quotient rule states that for a function \(\frac{f(x)}{g(x)}\), the derivative is: \(\frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\) Applying the quotient rule: \( R'(x) = \frac{(50)(0.01x^2 + 1) - (50x)(0.02x)}{(0.01x^2 + 1)^2}\) Now, we set R'(x) = 0 and solve for x: \( \frac{(50)(0.01x^2 + 1) - (50x)(0.02x)}{(0.01x^2 + 1)^2} = 0\) Since the denominator doesn't affect the equation equaling zero, we only need to consider the numerator: \( (50)(0.01x^2 + 1) - (50x)(0.02x)= 0 \)

Step 3: Solve for the optimal quantity demanded (x)

Now, let's solve for x in the equation obtained in step 2: \( 50(0.01x^2 + 1) - (0.02x)(50x) = 0\) Simplifying: \( 0.5x^2 + 50 - 1x^2 = 0\) \( -0.5x^2 + 50 = 0\) Dividing both sides by -0.5: \( x^2 - 100 = 0\) Factoring the quadratic equation: \((x + 10)(x - 10) = 0\) Possible solutions for x: \(x = 10\) or \(x = -10\) Since the quantity demanded cannot be negative, x = 10 is the optimal quantity demanded in thousands. To find the quantity demanded in units, we just multiply by 1000: \(x = 10 \cdot 1000 = 10000\)

Step 4: Answer

To yield maximum revenue, 10,000 watches must be sold.

Key Concepts

Demand Function

In economics, the demand function is a crucial concept that relates the price of a commodity to the quantity of the commodity that consumers are willing and able to purchase. A well-crafted demand function can reveal consumer behavior patterns and how various factors like price changes affect demand.

For the Sicard wristwatch, the demand function is given by the formula:

• \( p = \frac{50}{0.01x^2 + 1} \)

This tells us the price \( p \) for a quantity \( x \) of watches, where \( x \) is measured in thousands. The demand function is an inverse relationship, meaning as the quantity \( x \) increases, the price \( p \) tends to decrease.

Understanding this relationship helps in determining how pricing strategies may affect sales, managing inventory based on demand expectations, and ultimately optimizing revenue.

Derivative Calculation

Derivative calculation is a fundamental concept in calculus used to find how a function changes at any given point. In this context, it helps determine how revenue changes with respect to the quantity of watches sold.

To find the critical points of a revenue function, we begin by calculating its derivative. For the given revenue function

• \( R(x) = \frac{50x}{0.01x^2 + 1} \)

we use the quotient rule for derivatives. The quotient rule states that the derivative of \( \frac{f(x)}{g(x)} \) is:

• \( \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \)

By applying this rule, we derive the function \( R'(x) \) and then find x-values where this derivative equals zero. These values are critical points, possible candidates for maximizing or minimizing revenue.

Critical Points

Critical points in mathematics are values where the derivative of a function is zero or undefined, often indicating potential maxima, minima, or saddle points.

For the revenue function of the Sicard wristwatch:

• \( R'(x) = 0 \)

Setting the derivative equal to zero gives us possible critical points, where the function might achieve maximum revenue. Solving \( \frac{(50)(0.01x^2 + 1) - (50x)(0.02x)}{(0.01x^2 + 1)^2} = 0 \) involves simplifying the expression to:

• \( 0.5x^2 + 50 - x^2 = 0 \)

Finding roots of this equation reveals the quantity of watches that potentially maximizes revenue.

Quadratic Equation

Quadratic equations are polynomial equations of degree two, typically in the form \( ax^2 + bx + c = 0 \). Solving them involves finding values for \( x \) that satisfy the equation.

In our Sicard wristwatch problem, the simplified quadratic equation is:

• \( x^2 - 100 = 0 \)

To solve it, we factor as follows:

• \( (x + 10)(x - 10) = 0 \)

This yields two solutions: \( x = 10 \) and \( x = -10 \). However, since negative demand makes no economic sense, we choose \( x = 10 \), indicating 10,000 units when realized in practical terms. Quadratic equations like these often underpin demand assessments and pricing strategies.

Quantity Demanded

Quantity demanded refers to the total amount of a product consumers are willing to purchase at a given price. It is a fundamental aspect of supply and demand models, which forecast how varying prices will influence purchasing quantities.

In our specific example of the Sicard wristwatch, by solving the quadratic equation:

• \( x^2 - 100 = 0 \)

We derive that the optimal quantity demanded is \( x = 10 \). Transforming this to real-world units, given that \( x \) is in thousands, means 10,000 watches must be sold to optimize revenue.

This analysis helps businesses determine optimal production levels and pricing, ensuring supply meets consumer demand efficiently, maximizing potential profit.