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Optimization Calculator

Find the vertex (maximum or minimum) of a quadratic function f(x) = ax² + bx + c. Use this when solving real-world optimization problems like maximizing profit or minimizing cost.

Last updated: May 2026

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About this calculator

A quadratic function f(x) = ax² + bx + c has exactly one vertex, which is either a global maximum (when a < 0) or a global minimum (when a > 0). The x-coordinate of the vertex is found using the vertex formula: x = -b / (2a). Substituting this back into the original function gives the optimal value: f(-b/2a) = a(-b/2a)² + b(-b/2a) + c. This result comes from completing the square or setting the first derivative equal to zero. When a > 0 the parabola opens upward and the vertex is a minimum; when a < 0 it opens downward and the vertex is a maximum. This principle underlies countless applied problems in economics, physics, and engineering.

How to use

For a fixed perimeter, a rectangle encloses the most area when it is a square. With perimeter = 100 units, each side = 100 / 4 = 25 units, so the optimal area = 25² = 625 square units — exactly what (perimeter / 4)² returns. For a perimeter of 60 units: (60 / 4)² = 15² = 225 square units. The square is always the area-maximizing rectangle for a given perimeter.

Frequently asked questions

How do you find the maximum or minimum of a quadratic function?

The maximum or minimum of f(x) = ax² + bx + c occurs at the vertex. First compute the x-coordinate with x = -b / (2a). Then substitute that value back into the function to get the optimal output. If a > 0 the result is a minimum; if a < 0 it is a maximum.

What is the difference between a maximum and minimum in quadratic optimization?

Whether the vertex is a maximum or minimum depends entirely on the sign of the leading coefficient a. When a > 0, the parabola opens upward and the vertex is the lowest point, a global minimum. When a < 0, the parabola opens downward and the vertex is the highest point, a global maximum. There is no other local extremum for a quadratic.

When should you use quadratic optimization in real-world problems?

Quadratic optimization applies whenever a relationship between two quantities follows a parabolic curve. Common examples include maximizing revenue (price × quantity where demand is linear), minimizing material cost for a fixed volume, and finding the peak height of a projectile. Any time a model has the form ax² + bx + c and you need the best possible value, the vertex formula gives the answer directly without calculus.