Problem 46 - Essential Calculus Early Transcendentals

Question

During the summer months Terry makes and sclls necklaces on the beach. Last summer he sold the necklaces for \(\$ 10\) each and his sales averaged 20 per day. When he increased the price by \(\$ 1,\) he found that the average decreased by two sales per day. (a) Find the demand function, assuming that it is linear. (b) If the material for each necklace costs Terry \(\$ 6,\) what should the selling price be to maximize his profit?

Step-by-Step Solution

Answer

(a) The demand function is \( q = -2p + 40 \). (b) The price should be \( \$13 \) to maximize profit.

Step 1: Identify the Key Variables

Let's define the key variables for our problem:- Let \( p \) be the price of one necklace.- Let \( q \) be the quantity sold per day.From the problem statement, initially, when the price \( p = 10 \) dollars, the quantity sold \( q = 20 \). When the price is increased by \( 1 \) dollar (i.e., \( p = 11 \)), the quantity sold decreases by \( 2 \) units (i.e., \( q = 18 \)).

Step 2: Establish Two Points

We have two points that relate price \( p \) and quantity \( q \):- Point 1: \( (10, 20) \)- Point 2: \( (11, 18) \)These points show how price and quantity are related linearly.

Step 3: Calculate the Slope of the Demand Function

The demand function is linear, which means it is of the form \( q = mp + b \), where \( m \) is the slope.The slope \( m \) is calculated as follows:\[ m = \frac{q_2 - q_1}{p_2 - p_1} = \frac{18 - 20}{11 - 10} = \frac{-2}{1} = -2 \]

Step 4: Write the Equation of the Line

With the slope \( m = -2 \), use point 1 \((10, 20)\) to find the demand function.Use the point-slope form \( q - q_1 = m(p - p_1) \):\[ q - 20 = -2(p - 10) \]Expand and simplify:\[ q = -2p + 20 + 20 \q = -2p + 40 \]

Step 5: Calculate the Revenue Function

Revenue is calculated as \( R = p \times q \). Substituting for \( q \) from the demand function:\[ R = p(-2p + 40) \R = -2p^2 + 40p \]

Step 6: Set Up the Profit Function

Profit \( P \) is given by subtracting costs from revenue. If the cost per necklace is \( 6 \), the total cost for \( q \) necklaces is \( 6q \). Therefore, the profit function is:\[ P = R - 6q \P = (-2p^2 + 40p) - 6(-2p + 40) \P = -2p^2 + 40p - (-12p + 240) \P = -2p^2 + 40p + 12p - 240 \P = -2p^2 + 52p - 240 \]

Step 7: Find the Price that Maximizes Profit

The maximum profit occurs at the vertex of the parabola described by the quadratic formula \( P = -2p^2 + 52p - 240 \).The vertex formula for a parabola \( ap^2 + bp + c \) is given by \( p = -\frac{b}{2a} \):\[ p = -\frac{52}{2(-2)} = \frac{52}{4} = 13 \]Thus, the selling price to maximize profit is \( \$ 13 \).

Key Concepts

Demand Function

Understanding the Demand Function is essential in predicting how price changes affect sales. A demand function shows the relationship between the price of a product and the quantity demanded by consumers. In Terry's case, we observed:

• Price of \(10 resulted in 20 sales per day.
• Price increase to \)11 reduced sales to 18 per day.

This constant rate of change between price and sales indicates a linear demand function. To express this, we can use the equation of a line: \[ q = mp + b \]where \( q \) represents quantity sold, \( p \) is the price, \( m \) is the slope, and \( b \) is the y-intercept. By calculating the slope \( m \) from the points (10, 20) and (11, 18), we determine the demand function as \( q = -2p + 40 \). This tells us that for every $1 increase in price, sales decrease by 2 units.

Profit Maximization

Profit maximization involves setting a price that maximizes the difference between revenue and costs. Here, Terry calculates profits by determining revenue (price times quantity) and subtracting the cost from it. For each necklace, the cost is $6.The profit function can be described as:\[ P = R - ext{Cost} = p(-2p + 40) - 6(-2p + 40) \]After simplification, we obtain:\[ P = -2p^2 + 52p - 240 \]To find the maximum profit, we need to determine the price \( p \) that gives the highest result for this function. This revolves around finding the vertex of the parabola (since the function is quadratic) to pinpoint that price.

Linear Relationship

A Linear Relationship identifies a straight-line correlation between two variables, such as price and quantity in Terry's problem. By using two data points, we can calculate a slope that indicates how changes in one variable (price) will affect the other (quantity).The formula for the slope \( m \) is:\[ m = \frac{q_2 - q_1}{p_2 - p_1} \]For Terry:

• Initial point: (10, 20)
• Adjusted point: (11, 18)
• Slope \( m = \frac{18 - 20}{11 - 10} = -2 \)

This slope reveals that sales decrease by 2 units for every $1 increase in price, clearly illustrating the linear relationship in his demand function \( q = -2p + 40 \).

Quadratic Formula

The Quadratic Formula is crucial for solving quadratic equations, which appear in the profit maximization problem to find the price that leads to maximum profit. A quadratic function has the form:\[ P = ap^2 + bp + c \]Terry's problem requires identifying the vertex of this quadratic to locate the maximum profit point, using:\[ p = -\frac{b}{2a} \]In Terry's function,

• \( a = -2 \), \( b = 52 \)

Calculating gives:\[ p = -\frac{52}{2(-2)} = 13 \]This formula helps us understand that the optimal selling price for Terry's necklaces, to achieve maximum profit, is $13.