Math Problem Statement
Solution
Let's solve the problem step-by-step.
Problem Summary:
-
Initial Conditions:
- At $8 per ticket, the theater can sell all 1700 seats.
- For every additional $1 increase in ticket price, 95 fewer tickets are sold.
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Questions:
- Part (a): Find the ticket price that maximizes revenue.
- Part (b): Find the number of seats sold at that price.
Solution:
Let:
- = number of $1 increases above the base price of $8.
- Ticket price, .
- Tickets sold, .
Revenue, , is given by:
Expanding and simplifying :
This is a quadratic function in terms of , with a negative leading coefficient, so it opens downward, meaning it has a maximum point.
To find the maximum, we use the vertex formula for a parabola, .
Calculations:
For :
Rounding to two decimal places, .
Substitute to Find Price and Seats:
-
Ticket Price:
So, the ticket price that maximizes revenue is approximately $12.95.
-
Number of Seats Sold:
Rounding to the nearest whole number, 1230 seats are sold at this price.
Answers:
- (a) Ticket price = $12.95
- (b) Number of seats sold = 1230
Would you like more details on any step or have additional questions?
Related Questions:
- How do we determine if a function opens upwards or downwards in quadratic equations?
- Why is the revenue maximized at the vertex of the parabola?
- What happens to the revenue if the price is increased beyond $12.95?
- How does the ticket price affect the demand in real-life scenarios?
- Can we solve this problem using calculus methods, and how?
Tip:
In problems involving maximization with a quadratic equation, always check the sign of the leading coefficient—negative means a maximum, while positive means a minimum.
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Math Problem Analysis
Mathematical Concepts
Quadratic Functions
Revenue Maximization
Vertex of a Parabola
Formulas
Revenue = (Ticket Price) x (Number of Tickets Sold)
Quadratic Vertex Formula: x = -b / (2a)
Theorems
Vertex of a Parabola
Suitable Grade Level
Grades 10-12
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