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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?

Short Answer

Expert verified
The demand function is \(q = -8p + 700\). Maximum revenue occurs at price \$43.75, yielding \$15,312.50.

Step by step solution

01

Understand the Information Given

We have two data points for the demand: when the price \(p = 25\), the demand \(q = 500\), and when the price \(p = 30\), the demand \(q = 460\). We need to find the linear demand function \(q(p)\).
02

Find the Slope of the Demand Curve

The slope \(m\) of the demand curve \(q(p)\) can be calculated using the formula for the slope of a line, \(m = \frac{\Delta y}{\Delta x}\). Here, \(\Delta q = 460 - 500 = -40\) and \(\Delta p = 30 - 25 = 5\), so \(m = \frac{-40}{5} = -8\).
03

Formulate the Demand Equation

Knowing that the general linear equation is \(q = mp + b\), we substitute \(m = -8\). Selecting one of the points, say \((25, 500)\), we substitute into the equation to find \(b\):\[ 500 = -8 \times 25 + b \Longrightarrow b = 700. \]Thus, the demand equation is \(q = -8p + 700\).
04

Write Revenue as a Function of Price

Revenue \(R(p)\) can be calculated as price \(p\) times demand \(q\), i.e., \(R(p) = p \times q(p)\). Substituting the demand function we found: \[ R(p) = p(-8p + 700) = -8p^2 + 700p. \]
05

Determine the Price that Maximizes Revenue

To maximize the revenue function \(R(p) = -8p^2 + 700p\), find its vertex. The vertex form for \(-ax^2 + bx\) is at \(p = \frac{-b}{2a}\). Here, \(a = -8\) and \(b = 700\), so:\[ p = \frac{-700}{2(-8)} = \frac{700}{16} = 43.75. \]
06

Calculate the Maximum Revenue

Substitute \(p = 43.75\) back into the revenue equation: \[ R(43.75) = -8(43.75)^2 + 700 \times 43.75 = 15312.5. \]
07

Graph the Revenue Function

Graphing \(R(p) = -8p^2 + 700p\) shows a downward parabola, peaking at \(p = 43.75\) with a maximum revenue of 15312.5. The maximum point of the graph corresponds to the vertex of the parabola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Revenue Maximization
Revenue maximization occurs when a business sets its price point where the marginal gain from selling one more unit equals zero.
This means finding the optimal price that generates the highest possible revenue. To achieve this, we first need to establish a revenue function, as we've done by multiplying the price by the demand.
In our example, the revenue function was found to be \( R(p) = -8p^2 + 700p \), which is a quadratic equation represented as a downward-opening parabola.
Key steps to find maximum revenue:
  • Identify the revenue expression from the established demand function, \( q(p) \).
  • Use the vertex formula, which for a quadratic \( -ax^2 + bx \) is \( p = \frac{-b}{2a} \), to calculate the price that maximizes revenue.
  • Substitute this price back into the revenue function to determine the precise maximum revenue.
Applying these methods to our quadratic revenue function helped us to find the price at \( 43.75 \) and a maximum revenue of \( 15312.5 \). This systematic process simplifies the task of determining the optimal price point.
Graphing Parabolas
A parabola is a common curve used in business to model various functions, including demand and revenue equations.
In our case, the revenue function \( R(p) = -8p^2 + 700p \) forms a downward parabola because of the negative coefficient of \( p^2 \).
To graph a parabola effectively:
  • Determine the direction it opens by checking the coefficient of the squared term. Negative means it opens downwards.
  • Find the vertex, which is the highest or lowest point. For revenue functions, the vertex shows the price and the corresponding maximum revenue.
  • Plot additional points for different price values to check the parabola's path.
Graphing helps in visually identifying the peak, which corresponds to maximum revenue in the case of a revenue function.
This visual representation is crucial in business for analyzing how changes in price affect total revenue.
Calculus in Business
Calculus is an invaluable tool for solving complex business problems by analyzing changes and optimizing results.
In this scenario, we used calculus principles to identify the revenue maximization point.
Here’s how calculus applies:
  • We leveraged the concept of differentiation to understand how revenue changes with price. Though not explicitly calculated, the idea of a derivative helps to determine slopes and maximum points efficiently.
  • By setting the derivative of the revenue function to zero, one can find critical points, which would then be evaluated to confirm maxima or minima.
  • The vertex formula from algebra aligns closely with these calculus principles, providing a straightforward way to find maximum values within a quadratic revenue model.
These calculations and interpretations allow businesses to predict and strategize with precision, ensuring competitive advantage and profitability.

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