Maximizing Revenue: Price & Attendance Trade-off

What This Problem Teaches

• How to model revenue as the product of price and quantity when both depend on the same variable
• Writing quadratic functions from real-world constraints and relationships
• Finding maximum values using the vertex formula for parabolas
• Understanding the economic concept of price elasticity through mathematical optimization
• Interpreting why revenue functions create downward-opening parabolas in competitive markets

Solution: Method 1 — The Revenue Optimization Approach

Step 1 — Express price and attendance in terms of x

Let x represent the price increase in dollars. The current admission price is P dollars, so after the increase, the new price becomes:

Currently, 2000 tickets are sold per day. For each $1 increase, 100 fewer people visit. So after an increase of x dollars, the number of tickets sold becomes:

Step 2 — Write the revenue function

Revenue equals price times quantity sold. Multiplying our expressions from Step 1:

Step 3 — Expand to standard quadratic form

Let's distribute this product to see the quadratic structure clearly:

Step 4 — Find the vertex for maximum revenue

This is a quadratic function in the form ax² + bx + c where:

• a = -100 (negative, so parabola opens downward)
• b = 2000 - 100P
• c = 2000P

The maximum occurs at the vertex, where x = -b/(2a):

Step 5 — Calculate the optimal admission price

The optimal price increase is x = 10 - 0.5P. The optimal total admission price is:

Solution: Method 2 — The Symmetry Approach

Step 1 — Identify the linear factors

Revenue is the product of two linear functions: price f(x) = P + x and attendance g(x) = 2000 - 100x. When we multiply two linear functions, we get a quadratic with specific symmetry properties.

Step 2 — Find where revenue equals zero

The revenue function R(x) = (P + x)(2000 - 100x) equals zero when either factor equals zero:

Step 3 — Use symmetry to find the maximum

For a quadratic function, the vertex (maximum) occurs exactly halfway between the x-intercepts. The x-intercepts are at x = -P and x = 20, so:

Step 4 — Calculate optimal price

The optimal admission price is P + x = P + (10 - 0.5P) = 10 + 0.5P, confirming our first method.

Verification

Let's verify our formula works by testing with a specific value. Suppose the current price is P = $20:

This means if the current price is already $20, no change is needed for maximum revenue. Let's check by substituting into our revenue function:

This function is clearly maximized when x = 0, confirming our result. ✓

Let's also verify with P = $10:

At x = 5: price = $15, attendance = 1500, revenue = $22,500
At x = 4: price = $14, attendance = 1600, revenue = $22,400
At x = 6: price = $16, attendance = 1400, revenue = $22,400

The revenue is indeed highest at our calculated optimum. ✓

Watch Out For These

The Math Beneath the Surface

This problem exemplifies a fundamental pattern in economics and optimization: revenue maximization with linear demand. The general form is:

When expanded, this always creates a quadratic function R(x) = ax² + bx + c where a < 0 (downward parabola). The coefficient a represents the demand sensitivity—how much quantity drops per unit price increase.

In our case, the sensitivity is 100 customers per dollar, giving a = -100. Higher sensitivity (steeper demand curve) means a sharper revenue peak, while lower sensitivity creates a flatter optimum.

The key insight: optimal pricing always balances the trade-off between margin and volume. Raise prices too much, and you lose too many customers. Keep prices too low, and you miss profit potential. The mathematical optimum finds that perfect balance.

Real Applications

Concert venues and sports teams use this exact model when setting ticket prices. They survey price sensitivity and optimize for maximum gate revenue, often finding that moderate price increases boost total income despite smaller crowds.

Subscription services like Netflix apply similar analysis when adjusting monthly fees. They measure subscriber churn rates against price increases to find the revenue-maximizing price point.

Manufacturing companies use revenue optimization when launching products. Higher prices mean better profit margins but fewer units sold. The optimal price maximizes total revenue before considering production costs.