Finding the Maximum or Minimum Value by Completing the Square or by Differentiation

Introduction

Key Concepts

Quadratic Functions and Their Graphs

A quadratic function is a second-degree polynomial of the form: $$ f(x) = ax^2 + bx + c $$ where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \). The graph of a quadratic function is a parabola that opens upwards if \( a > 0 \) and downwards if \( a < 0 \). The highest or lowest point on the parabola is known as the vertex, which represents the maximum or minimum value of the function.

Vertex Form of a Quadratic Function

The vertex form of a quadratic function provides a straightforward way to identify the vertex of the parabola: $$ f(x) = a(x - h)^2 + k $$ where \( (h, k) \) is the vertex of the parabola. Converting a quadratic function from standard form to vertex form involves completing the square, enabling easy identification of the function's maximum or minimum value.

Completing the Square

Completing the square is a method used to rewrite a quadratic equation in vertex form. It involves manipulating the equation to form a perfect square trinomial, which allows for the identification of the vertex. The general steps are:

1. Start with the standard form: \( f(x) = ax^2 + bx + c \).
2. Factor out the coefficient of \( x^2 \) if \( a \neq 1 \): \( f(x) = a(x^2 + \frac{b}{a}x) + c \).
3. Add and subtract \(\left(\frac{b}{2a}\right)^2\) inside the parentheses to complete the square:
$$ f(x) = a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c $$
4. Simplify to obtain the vertex form: \( f(x) = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) \).

This reveals that the vertex is at \( \left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right) \), providing the maximum or minimum value depending on the direction the parabola opens.

Differentiation

Differentiation is a powerful tool in calculus used to find the rate at which a function is changing at any given point. For quadratic functions, differentiation can be employed to determine the critical points, which correspond to the function's maximum or minimum values.

Given a quadratic function \( f(x) = ax^2 + bx + c \), the derivative \( f'(x) \) is: $$ f'(x) = 2ax + b $$ Setting the derivative equal to zero to find critical points: $$ 2ax + b = 0 \\ x = -\frac{b}{2a} $$ Substituting this value back into the original function provides the y-coordinate of the vertex: $$ f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c = c - \frac{b^2}{4a} $$

Maximum and Minimum Values

The maximum or minimum value of a quadratic function is determined by the vertex of its parabola. If \( a > 0 \), the parabola opens upwards, and the vertex represents the minimum value. Conversely, if \( a < 0 \), the parabola opens downwards, and the vertex signifies the maximum value.

Using the vertex form or differentiation provides a systematic approach to finding these extremum points, which are crucial in various applications such as optimization problems.

Applications of Maximum and Minimum Values

Identifying the maximum or minimum values of quadratic functions has diverse applications:

• Physics: Calculating the optimal speed or trajectory for projectiles to achieve maximum distance.
• Economics: Determining cost functions' minima to maximize profit.
• Engineering: Designing structures that require optimization for material usage and stress distribution.
• Biology: Modeling population growth and determining carrying capacities.

Examples

Let's explore some examples to solidify the understanding of finding maximum or minimum values using both completing the square and differentiation.

Example 1: Using Completing the Square

Find the minimum value of the function \( f(x) = 2x^2 - 8x + 5 \).

Solution:

1. Start with the standard form: $$ f(x) = 2x^2 - 8x + 5 $$
2. Factor out the coefficient of \( x^2 \): $$ f(x) = 2(x^2 - 4x) + 5 $$
3. Complete the square inside the parentheses: $$ x^2 - 4x = x^2 - 4x + 4 - 4 = (x - 2)^2 - 4 $$
4. Substitute back into the equation: $$ f(x) = 2[(x - 2)^2 - 4] + 5 = 2(x - 2)^2 - 8 + 5 = 2(x - 2)^2 - 3 $$
5. Identify the vertex \( (h, k) = (2, -3) \).
6. Since \( a = 2 > 0 \), the function has a minimum value of \(-3\).

Example 2: Using Differentiation

Find the maximum value of the function \( f(x) = -3x^2 + 12x - 5 \).

Solution:

1. Find the derivative: $$ f'(x) = -6x + 12 $$
2. Set the derivative equal to zero to find critical points: $$ -6x + 12 = 0 \\ x = 2 $$
3. Substitute \( x = 2 \) back into the original function to find \( f(2) \): $$ f(2) = -3(2)^2 + 12(2) - 5 = -12 + 24 - 5 = 7 $$
4. Since \( a = -3 < 0 \), the function has a maximum value of \(7\).

Common Mistakes to Avoid

When finding maximum or minimum values, students often encounter the following pitfalls:

• Incorrectly completing the square by miscalculating coefficients.
• Forgetting to factor out the leading coefficient before completing the square.
• Errors in taking derivatives, especially sign mistakes.
• Misinterpreting the vertex coordinates, leading to incorrect maximum or minimum values.
• Not considering the direction of the parabola, which determines whether the vertex is a maximum or minimum.

Advanced Concepts

Mathematical Derivations and Proofs

Delving deeper into the theoretical underpinnings, we can derive the vertex form of a quadratic function using completing the square and explore its implications in calculus.

Derivation of Vertex Form

Given the standard form of a quadratic function: $$ f(x) = ax^2 + bx + c $$ To complete the square: \begin{align*} f(x) &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\ &= a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) \end{align*} Thus, the vertex form is: $$ f(x) = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) $$ which identifies the vertex at \( \left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right) \).

Second Derivative Test

In calculus, the second derivative test provides a method to determine the concavity of a function and thus classify critical points as maxima or minima.

Given the quadratic function \( f(x) = ax^2 + bx + c \), its first derivative is: $$ f'(x) = 2ax + b $$ The second derivative is: $$ f''(x) = 2a $$ Since \( f''(x) = 2a \) is constant:

• If \( a > 0 \), \( f''(x) > 0 \), indicating the function is concave upwards and has a minimum at the critical point.
• If \( a < 0 \), \( f''(x) < 0 \), indicating the function is concave downwards and has a maximum at the critical point.

Complex Problem-Solving

Applying these concepts to more complex scenarios enhances problem-solving proficiency. Consider optimization problems where multiple constraints are involved.

Optimizing Revenue

A company wants to determine the price \( p \) that maximizes its revenue. If the demand function is linear and given by \( D(p) = 100 - 2p \), where \( D(p) \) is the quantity demanded at price \( p \), the revenue \( R \) is: $$ R(p) = p \times D(p) = p(100 - 2p) = 100p - 2p^2 $$ To find the price that maximizes revenue:

1. Express the revenue function: $$ R(p) = -2p^2 + 100p $$
2. Find the derivative: $$ R'(p) = -4p + 100 $$
3. Set the derivative equal to zero: $$ -4p + 100 = 0 \\ p = 25 $$
4. Verify using the second derivative: $$ R''(p) = -4 < 0 $$ Thus, \( p = 25 \) yields a maximum revenue.
5. Maximum revenue: $$ R(25) = 100(25) - 2(25)^2 = 2500 - 1250 = 1250 $$

Therefore, setting the price at \$25 maximizes the company's revenue to \$1250.

Engineering Design Optimization

In engineering, optimizing material usage while maintaining structural integrity is paramount. For instance, determining the dimensions of a rectangular box with maximum volume for a given surface area involves quadratic relationships.

Let the dimensions of the box be length \( l \), width \( w \), and height \( h \). Given a fixed surface area \( S \), the volume \( V \) to be maximized is: $$ V = lwh $$ Using constraints and expressing variables in terms of each other leads to a quadratic optimization problem solvable by completing the square or differentiation.

Interdisciplinary Connections

The concepts of maximum and minimum values extend beyond pure mathematics, finding relevance in various disciplines:

• Physics: Optimization of forces, energy minimization in systems, and maximizing work done.
• Economics: Cost minimization and profit maximization strategies.
• Biology: Modeling population dynamics and resource optimization.
• Computer Science: Algorithm optimization for efficiency and resource management.
• Environmental Science: Optimizing resource use to balance ecological impact.

These interdisciplinary applications underscore the versatility and importance of mastering maximum and minimum value calculations in quadratic functions.

Real-World Applications

Exploring real-world scenarios where these mathematical concepts are applied enhances understanding and appreciation for their practical significance:

• Architecture: Designing structures that use materials efficiently while maintaining strength and aesthetics.
• Aerospace Engineering: Calculating optimal flight paths and fuel consumption for maximum efficiency.
• Finance: Portfolio optimization to maximize returns and minimize risks.
• Medicine: Modeling drug dosage levels to achieve maximum therapeutic effect with minimal side effects.
• Agriculture: Optimizing crop yields based on resource allocation and environmental factors.

These applications demonstrate the integral role of mathematical optimization in solving complex, real-life problems across various sectors.

Comparison Table

Summary and Key Takeaways