Math Problem Statement

The area of a rectangle is modeled by the graph where y is the area (in square meters) and x is the width (in meters). Find the maximum area of the rectangle.

Solution

The problem provided shows a graph that models the area yy of a rectangle as a function of its width xx, where yy is the area in square meters and xx is the width in meters.

Key observations:

  • The graph seems to be a parabola with a vertex, and it is symmetric, suggesting the equation might be quadratic.
  • The maximum area corresponds to the highest point on the graph, which is the vertex of the parabola.

Steps to solve:

  1. Analyze the graph:

    • From the graph, it looks like the parabola has a maximum at x=3x = 3 meters (the width) and a maximum area y=9y = 9 square meters (since the graph peaks around y=9y = 9).
  2. Conclusion:

    • The maximum area of the rectangle is 9 square meters, and this occurs when the width x=3x = 3 meters.

Would you like to dive into the mathematical reasoning behind quadratic equations and parabolas? Or do you have more questions about this?

Relative questions:

  1. What is the general form of a quadratic equation and how does it relate to the graph?
  2. Can you explain how the vertex form of a parabola helps find the maximum value?
  3. How can the width and height of the rectangle be found from the graph?
  4. Why does the maximum area occur at the vertex of the parabola?
  5. How would the equation change if the width were constrained to a different range?

Tip:

Always check the vertex of a parabola when searching for maximum or minimum values, as it directly gives the peak (maximum) or trough (minimum).

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Parabolas

Formulas

Vertex form of a quadratic equation: y = ax^2 + bx + c
Maximum value of a parabola at the vertex: x = -b/(2a)

Theorems

Vertex theorem for parabolas

Suitable Grade Level

Grades 8-10