Quadratic Optimization

Quadratic optimization is the process of finding the maximum or minimum value of a quadratic function — a parabola — without using calculus. Every quadratic function of the form f(x) = ax² + bx + c has exactly one extreme value: a minimum if the parabola opens upward (a > 0) or a maximum if it opens downward (a < 0). This extreme value occurs at the vertex of the parabola, making vertex-finding the central skill in quadratic optimization. Although this topic appears in the study of equations and inequalities, its applications reach far beyond algebra. Businesses use it to maximize profit or minimize cost; engineers use it to design projectile paths and structural arches; even athletes and game designers rely on the same mathematics to model real trajectories. The power of quadratic optimization is that a simple algebraic technique — completing the square or using the vertex formula — replaces the need for the more advanced tools of calculus. At the high school level, quadratic optimization problems typically ask you to identify what quantity is being optimized (maximized or minimized), express that quantity as a quadratic function of a single variable, locate the vertex, and interpret the result in context. Mastery of this concept ties together factoring, completing the square, the quadratic formula, and graphing into one unified problem-solving strategy.

Key takeaways
  • Recognize the Standard Form
  • Use the Vertex Formula
  • Complete the Square (Alternative Method)
  • Set Up the Objective Function
  • Interpret Within Context and Check Domain

Recognize the Standard Form

A quadratic function is written f(x) = ax² + bx + c, where a ≠ 0. The sign of a determines whether the parabola opens upward (a > 0, giving a minimum) or downward (a < 0, giving a maximum). Always identify a first, because this tells you whether you are looking for a minimum or a maximum.

Use the Vertex Formula

The x-coordinate of the vertex is given by x = -b/(2a). This single formula locates the input value at which the extreme value occurs. Substitute this x-value back into f(x) to get the optimal output value: f(-b/(2a)). This two-step calculation is the fastest algebraic route to the answer.

Complete the Square (Alternative Method)

Rewriting f(x) = ax² + bx + c in vertex form f(x) = a(x - h)² + k reveals the vertex (h, k) directly. To complete the square: factor a from the first two terms, add and subtract (b/2a)² inside the bracket, then simplify. For example, f(x) = 2x² - 8x + 3 becomes 2(x - 2)² - 5, so the vertex is (2, -5) and the minimum value is -5.

Set Up the Objective Function

In word problems, you must first build the quadratic. Identify the quantity to optimize (area, revenue, height, etc.) and express it using a single variable. Use any constraint given — such as a fixed perimeter or total quantity — to eliminate a second variable. The result should be a quadratic function in one variable ready for vertex analysis.

Interpret Within Context and Check Domain

The mathematical vertex gives the optimal value, but you must check that it lies within a physically or contextually valid domain. For example, a length cannot be negative. If the vertex falls outside the valid domain, then the optimum occurs at a boundary value instead. Always state your answer in the units and language of the original problem.

The Key Equation

xvertex=−b2a,f ⁣(−b2a)=c−b24ax_{\text{vertex}} = -\frac{b}{2a}, \qquad f\!\left(-\frac{b}{2a}\right) = c - \frac{b^2}{4a}
  • a — Leading coefficient (dimensionless): Determines the direction and steepness of the parabola. If a > 0 the vertex is a minimum; if a < 0 it is a maximum.
  • b — Linear coefficient (dimensionless): Shifts the axis of symmetry left or right. Together with a it sets the location of the vertex on the x-axis.
  • c — Constant term (dimensionless): The y-intercept of the parabola. It shifts the entire parabola up or down without changing the x-coordinate of the vertex.
  • x_{\text{vertex}} — Optimal input (same as x): The value of the independent variable that produces the maximum or minimum output.

Worked Example

Problem. A farmer has 120 metres of fencing to enclose a rectangular field. One side of the rectangle lies along a river and needs no fencing. What dimensions maximize the area, and what is the maximum area?

y=30 m,x=60 m,Amax⁡=1800 m2y = 30\text{ m},\quad x = 60\text{ m},\quad A_{\max} = 1800\text{ m}^2

Answer. The field should be 60 m along the river and 30 m wide, giving a maximum area of 1800 square metres.

Figure
Diagram
Vertex (30, 1800)MaximumyA0153045601800900Width y (metres)Area A (m²)A(y) = −2y² + 120y
Origins · 1 minPierre de Fermat and how this was discovered
PD
Pierre de Fermat
1607–1665 · French

Fermat developed a method called 'adequality' (pseudo-equality) around 1629 to find the maximum and minimum values of polynomial functions — well before calculus was formally invented by Newton and Leibniz. His technique set x equal to x + e, expanded the expression, and then set the leading term in e to zero, effectively pre-inventing differentiation. For quadratics, his method reproduces exactly the vertex formula x = −b/(2a). Fermat also applied optimization to problems in optics, deriving the law of refraction (Snell's law) from the principle that light travels the path of least time. Fermat was a lawyer by profession and did mathematics entirely as a hobby, yet he made foundational contributions to number theory, analytic geometry, probability, and calculus — often scribbling results in margins of books without publishing proofs.

  • The Quadratic Formula — The vertex x-coordinate x = −b/(2a) is derived from the quadratic formula; understanding both together deepens algebraic fluency.
  • Completing the Square — Completing the square converts a quadratic to vertex form, the most direct algebraic route to reading off the maximum or minimum value.
  • Linear Inequalities — Domain restrictions in optimization problems are often expressed as linear inequalities that constrain which vertex values are physically meaningful.
  • Parabolas and Quadratic Graphs — Graphing the parabola gives visual confirmation of whether a vertex is a maximum or minimum and helps check that optimization answers are reasonable.
  • Introduction to Derivatives — Calculus generalizes quadratic optimization to any differentiable function; the derivative equals zero exactly at the vertex, connecting algebra to calculus.