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You have been hired as a marketing consultant to Johannesburg Burger Supply, Inc., and you wish to come up with a unit price for its hamburgers in order to maximize its weekly revenue. To make life as simple as possible, you assume that the demand equation for Johannesburg hamburgers has the linear form \(q=m p+b\), where \(p\) is the price per hamburger, \(q\) is the demand in weekly sales, and \(m\) and \(b\) are certain constants you must determine. a. Your market studies reveal the following sales figures: When the price is set at \(\$ 2.00\) per hamburger, the sales amount to 3,000 per week, but when the price is set at \(\$ 4.00\) per hamburger, the sales drop to zero. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes weekly revenue and predict what the weekly revenue will be at that price.

Step by step solution

01

Find the slope (m) using given data points

Let's use the point-slope form to find the equation of the line. We know two points on this line: \((2, 3000)\) and \((4, 0)\) The slope \(m\) is defined as: \[m = \frac{y_2 - y_1}{x_2 - x_1}\] Substituting the given points: \[m = \frac{0 - 3000}{4 - 2} = \frac{-3000}{2} = -1500\] So, the slope \(m = -1500\).

02

Find the constant (b) using one of the points and the slope

To find the constant \(b\), let's use one of the given points and the slope we calculated in Step 1. We can use the point \((2, 3000)\): \(q = mp + b\) Substituting the values: \[3000 = -1500 \times 2 + b\] \[b = 3000 + 3000 = 6000\] Now we have the demand equation: \[q = -1500p + 6000\]

03

Find the revenue equation

The revenue equation is given by: \[R = pq\] Substituting the demand equation we found earlier: \[R(p) = p(-1500p + 6000)\] To make it more readable, let's simplify the equation: \[R(p) = -1500p^2 + 6000p\]

04

Find the price that maximizes the revenue

To find the maximum revenue, we will find the derivative of the revenue function and set it equal to zero. \[\frac{dR}{dp} = \frac{d}{dp}(-1500p^2 + 6000p)\] Using the power rule of differentiation, we get: \[\frac{dR}{dp} = -3000p + 6000\] Now, let's set the derivative to zero and solve for \(p\): \[-3000p + 6000 = 0\] \[p = \frac{6000}{3000} = 2\] Hence, the price that maximizes the revenue is \(\$2.00\).

05

Calculate the weekly revenue at the maximum price

To find the weekly revenue at the optimal price, we need to substitute the price in the revenue equation: \[R(2) = -1500(2)^2 + 6000(2)\] \[R(2) = -6000 + 12000 = 6000\] Therefore, the maximum weekly revenue is \(\$6000\).

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Key Concepts

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

Linear Demand Equation

In the realm of economics, a Linear Demand Equation expresses the relationship between the price of a product and the quantity demanded by consumers. When plotted on a graph, it forms a straight line. The equation is typically represented by the formula:

• \[ q = mp + b \]

where:

• \( q \) = quantity demanded
• \( p \) = price per unit
• \( m \) = slope of the demand curve
• \( b \) = y-intercept, representing demand when price is zero

In our exercise, the linear demand equation helps determine how many hamburgers will be sold at varying prices. We used two data points: when the price was \(\\(2.00\) and 3,000 hamburgers were sold, and when the price was \(\\)4.00\) and no hamburgers were sold. This information allowed us to calculate the demand equation as \( q = -1500p + 6000 \). Understanding this equation is crucial to optimizing pricing strategies.

Differentiation

Differentiation is a mathematical process used to determine how a function's output changes with respect to its inputs. It's essentially used to find the slope or rate of change at a given point on a curve. In revenue maximization problems, differentiation reveals how changes in price impact revenue.
To maximize revenue, we differentiate the revenue function with respect to price. For our revenue function:

• \[ R(p) = -1500p^2 + 6000p \]

we calculate the derivative as:

• \[ \frac{dR}{dp} = -3000p + 6000 \]

Setting the derivative equal to zero helps identify price points where revenue doesn't change, indicating a maximum or minimum. Solving:

• \[-3000p + 6000 = 0 \]
• \[ p = 2 \]

The solution \( p = \$2.00 \) indicates the price point where we expect the greatest revenue generation.

Slope Calculation

The slope of a line in a Linear Demand Equation represents the rate at which demand changes for each unit change in price. It is a crucial concept in understanding the responsiveness of demand to price changes.

To calculate the slope, we use the formula:

• \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Given the points \((2, 3000)\) and \((4, 0)\), we find the slope to be:

• \[ m = \frac{0 - 3000}{4 - 2} = -1500 \]

This slope indicates that for every dollar increase in price, the demand decreases by 1,500 units. It's essential to recognize that a steeper slope means more sensitivity to price changes. Understanding the slope helps businesses make informed decisions about pricing strategies to balance price with potential sales volume. It's a foundational aspect of crafting a competitive pricing policy.