Problem 29 - Applied Calculus

Question

A sporting goods wholesaler finds that when the price of a product is \(\$ 25,\) the company sells 500 units per week. When the price is \(\$ 30,\) the number sold per week decreases to 460 units. (a) Find the demand, \(q\), as a function of price, \(p\), assuming that the demand curve is linear. (b) Use your answer to part (a) to write revenue as a function of price. (c) Graph the revenue function in part (b). Find the price that maximizes revenue. What is the revenue at this price?

Step-by-Step Solution

Answer

(a) Demand is \( q = -8p + 700 \). (b) Revenue is \( R(p) = -8p^2 + 700p \). (c) Max revenue at \( p = 43.75\) is \( 15,312.5 \).

Step 1: Understand the Given Data

We're given two points on the demand curve: when the price \( p = 25 \), the demand \( q = 500 \); and when \( p = 30 \), the demand \( q = 460 \). We need to find the relationship between price and demand, assuming it's linear.

Step 2: Find the Slope of the Demand Curve

The formula for the slope of a line between two points \((p_1, q_1)\) and \((p_2, q_2)\) is given by \( m = \frac{q_2 - q_1}{p_2 - p_1} \). Substituting the known values, we have:\[ m = \frac{460 - 500}{30 - 25} = \frac{-40}{5} = -8 \].

Step 3: Write the Equation of the Demand Line

Using the point-slope form, \( q - q_1 = m(p - p_1) \), substitute \( m = -8 \) and \( (p_1, q_1) = (25, 500) \):\[ q - 500 = -8(p - 25) \]Simplify to get the equation of the demand line:\[ q = -8p + 700 \].

Step 4: Write the Revenue Function

Revenue \( R \) is the product of price \( p \) and quantity \( q \). From Step 3, \( q = -8p + 700 \). Thus, revenue is:\[ R(p) = p \cdot (-8p + 700) = -8p^2 + 700p \].

Step 5: Find the Price that Maximizes Revenue

The revenue function \( R(p) = -8p^2 + 700p \) is a quadratic function opening downwards (coefficient of \( p^2 \) is negative). The vertex of a parabola \( ax^2 + bx + c \) is at \( p = -\frac{b}{2a} \). Here, \( a = -8 \) and \( b = 700 \):\[ p = -\frac{700}{2(-8)} = \frac{700}{16} = 43.75 \].

Step 6: Find the Maximum Revenue

Substitute \( p = 43.75 \) into \( R(p) = -8p^2 + 700p \):\[ R(43.75) = -8(43.75)^2 + 700(43.75) \]. Calculate this to find the revenue:\[ R(43.75) = 15312.5 \].

Step 7: Graph Revenue Function and Confirm Maximization Price

Carefully plot the quadratic revenue function \( R(p) = -8p^2 + 700p \) and show that the maximum point (vertex) aligns with our calculated price of \( p = 43.75 \). This confirms that the revenue maximizes as calculated.

Key Concepts

The Concept of a Demand Curve

A demand curve illustrates the relationship between the price of a product and the quantity demanded by consumers. It is depicted graphically, often sloping downwards, which reflects the law of demand: as prices decrease, demand tends to increase, and vice versa. In our case, we are dealing with a **linear demand curve**, which essentially means it’s a straight line. This simplifies the relationship into a basic algebraic equation, usually written in the form of \( q = mp + c \).
Here, \( m \) is the slope of the demand curve, which indicates how demand changes with price, and \( c \) is the intercept, demonstrating demand when price is zero. In practical terms, a negative slope reflects that demand decreases as price increases. In our exercise, this is precisely what was calculated: the slope was found to be -8, meaning for every $1 increase in price, demand decreases by 8 units.
This linear approach makes calculating different price-demand scenarios straightforward, giving businesses a valuable tool in making pricing decisions.

Understanding the Revenue Function

The revenue function represents how a company's earnings change with different sales volumes and prices. It's defined as the product of price \( p \) and quantity demanded \( q \). Therefore, in symbolic terms, the revenue function is written as \( R(p) = p \, q(p) \), where \( q(p) \) denotes demand for a given price.
From the previous calculations, we know that \( q(p) = -8p + 700 \). By substituting this into the revenue function, we get \( R(p) = p(-8p + 700) \). After simplifying, the revenue function becomes \( R(p) = -8p^2 + 700p \). This quadratic equation graphically manifests as a parabola.
It's crucial to note that the coefficient of \( p^2 \) is negative, indicating the parabola opens downward. This characteristic is significant because it tells us there's a peak revenue point, which is where revenue is maximized.

Maximizing Revenue

Maximizing revenue involves identifying the price at which the company earns the most money. With a quadratic revenue function, the task is to find the **vertex** of the parabola, since it's the highest point on a downward-opening curve. Mathematically, the price that maximizes revenue in a quadratic function \( R(p) = ap^2 + bp + c \) can be found using the vertex formula \( p = -\frac{b}{2a} \).
Applying this formula, given that \( a = -8 \) and \( b = 700 \), we find the optimal price is \( p = 43.75 \). This means the sporting goods wholesaler will maximize its revenue by setting the price at \(43.75. Calculating further using this price in the revenue function, the maximum revenue can be established as approximately \)15,312.50.
Setting the right price is crucial for businesses aiming to maximize earnings. This calculated price reflects the perfect balance between price level and sales volume to achieve optimal financial performance.