Problem 22 - Applied Calculus

Question

An ice cream company finds that at a price of \(\$ 4.00\), demand is 4000 units. For every \(\$ 0.25\) decrease in price, demand increases by 200 units. Find the price and quantity sold that maximize revenue.

Step-by-Step Solution

Answer

Optimal price: \$3.50, demand: 4400 units.

Step 1: Understanding the Problem

To maximize revenue, we first need to express revenue as a function of price or demand. Revenue is defined as the product of price and quantity (demand).

Step 2: Define Variables and Relationships

Let the price decrease by \(x\) increments of \$0.25. Hence, the new price is \(4 - 0.25x\). The demand increases by 200 units for each decrement: 4000 + 200x.

Step 3: Write the Revenue Function

The revenue \(R\) as a function of \(x\) is given by \(R(x) = (4 - 0.25x) \times (4000 + 200x)\). This represents revenue in terms of price decreases and resulting demand.

Step 4: Expand the Revenue Function

Expand the function: \[ R(x) = (4)(4000) + (4)(200x) - (0.25x)(4000) - (0.25x)(200x) \] which simplifies to \[ R(x) = 16000 + 800x - 1000x - 50x^2 \] thus \[ R(x) = -50x^2 - 200x + 16000 \].

Step 5: Maximize the Revenue using Quadratic Formula

Since \(R(x)\) is a downward opening parabola, the maximum value occurs at its vertex. Find \(x_{max}\) using \(x = -\frac{b}{2a}\). Here, \(a = -50\), \(b = -200\).\[ x = -\frac{-200}{2 \times -50} = 2 \].

Step 6: Calculate Optimal Price and Demand

Substitute \(x = 2\) back into the formulas for price and demand: Price: \(4 - 0.25 \times 2 = \$ 3.50\) Demand: \(4000 + 200 \times 2 = 4400\).

Key Concepts

Demand Function

The demand function is a key concept in economics. It shows the relationship between the price of a good and the quantity demanded by consumers.
In our ice cream company example, the demand can be influenced by changes in price. Essentially, as the price decreases, more units are demanded, following the law of demand.

• Starting with a base price of \(4, the demand is 4000 units.
• For each \)0.25 decrease in price, demand increases by 200 units.

This relationship can be expressed as a linear equation. If the price is decreased by \(x\) increments of $0.25, the new demand is given by the formula: \(4000 + 200x\). This equation helps us track how demand fluctuates with price changes, providing crucial data for the firm's pricing strategy.

Price Optimization

Price optimization involves finding the ideal pricing point for a product, one that maximizes revenue. In our scenario, this means adjusting the price of ice cream to balance demand and price.
To achieve optimal pricing, we first express revenue as a function of price and demand. Revenue \( R \) is calculated as:

• \( R(x) = (Price) \times (Demand) \)
• For our exercise, \( R(x) = (4 - 0.25x) \times (4000 + 200x) \).

Next, we expand the equation using algebraic techniques. This gives us a clear equation to apply calculus methods for finding maxima. Price optimization requires not just setting prices but understanding the market’s response to price changes. The chosen price should yield the highest possible revenue without significantly reducing demand.

Quadratic Function

A quadratic function is a polynomial function of degree two, often used to model real-world phenomena like revenue maximization. The standard form is \( ax^2 + bx + c \).
For our revenue function, this takes the form \( R(x) = -50x^2 - 200x + 16000 \).

• The coefficient \(-50x^2\) shows it is a downward-opening parabola. This indicates that there is a maximum point for the revenue function, rather than a minimum.
• We find the maximum point, or the vertex, to determine the optimal number of price changes \( x \) that maximizes revenue.

Using the vertex formula \( x = -\frac{b}{2a} \), we calculate the location of this maximum. Here, \( a = -50 \),\( b = -200 \), leading to \( x = 2 \).
This point of \( x = 2 \) gives us both the optimal price and demand figures when plugged back into the demand function.