Applications of Derivatives: Extrema (Maxima and Minima)

Maxima and Minima: Introduction and Concepts

One of the most significant applications of derivatives is in finding the maximum and minimum values of a function. These extreme values are often referred to as extrema (plural of extremum).

Finding maxima and minima is crucial in optimization problems, where we aim to find the best possible outcome under certain constraints (e.g., maximizing profit, minimizing cost, maximizing area, minimizing distance).

Basic Idea and Definitions

The goal is to find the highest or lowest points on the graph of a function, either within a specific interval or over its entire domain.

• Maximum Value: A function $f$ has a maximum value at a point $c$ if $f(c)$ is the largest value that the function attains. This means $f(c) \ge f(x)$ for all $x$ in the specified domain or interval. The value $f(c)$ is the maximum value, and $c$ is the location where the maximum occurs.
• Minimum Value: A function $f$ has a minimum value at a point $c$ if $f(c)$ is the smallest value that the function attains. This means $f(c) \le f(x)$ for all $x$ in the specified domain or interval. The value $f(c)$ is the minimum value, and $c$ is the location where the minimum occurs.

Maximum and minimum values are collectively called extrema (the plural of extremum).

Local vs. Absolute Extrema

Extrema can be classified based on whether they represent the extreme values in a small neighborhood (local) or over the entire domain (absolute).

• Local Maximum (Relative Maximum): A function $f$ has a local maximum at $c$ if $f(c) \ge f(x)$ for all $x$ in some open interval containing $c$. This means $f(c)$ is the highest point in its immediate vicinity on the graph. It's a "peak" or the top of a "hill" on the curve.
• Local Minimum (Relative Minimum): A function $f$ has a local minimum at $c$ if $f(c) \le f(x)$ for all $x$ in some open interval containing $c$. This means $f(c)$ is the lowest point in its immediate vicinity. It's a "valley" or the bottom of a "dip" on the curve.
• Absolute Maximum (Global Maximum): A function $f$ has an absolute maximum at $c$ if $f(c) \ge f(x)$ for all $x$ in the entire domain of $f$. This is the overall highest value the function ever reaches across its whole domain.
• Absolute Minimum (Global Minimum): A function $f$ has an absolute minimum at $c$ if $f(c) \le f(x)$ for all $x$ in the entire domain of $f$. This is the overall lowest value the function ever reaches across its whole domain.

Local maxima and local minima are together called local extrema. Absolute maxima and absolute minima are together called absolute extrema or global extrema.

A function may have several local maxima and minima, but it can have at most one absolute maximum value and at most one absolute minimum value (though these values can be attained at more than one point). The absolute maximum (or minimum) may occur at a local maximum (or minimum) or at an endpoint of a closed interval.

Critical Points

Derivatives help us locate the possible points where local extrema can occur.

A critical point (or critical number) of a function $f$ is a number $c$ in the domain of $f$ such that either:

1. The first derivative is zero at $c$: $f'(c) = 0$. This corresponds to a point on the graph where the tangent line is horizontal.
2. The first derivative does not exist at $c$: $f'(c)$ is undefined. This can occur at points like sharp corners, cusps, or vertical tangents.

Fermat's Theorem (for Local Extrema): If $f$ has a local maximum or minimum at $c$, and if $f'(c)$ exists, then $f'(c) = 0$.

This theorem is fundamental: it tells us that if a local extremum occurs at a point where the function is differentiable, then the tangent line at that point must be horizontal ($f'(c)=0$). This means that points where $f'(c)=0$ are candidates for local extrema. However, the converse is not true; a point where $f'(c)=0$ is not necessarily a local extremum (e.g., $f(x)=x^3$ has $f'(0)=0$ at $x=0$, but $x=0$ is an inflection point, not an extremum).

Furthermore, local extrema can also occur at points where the derivative does not exist (e.g., the sharp corner of $f(x)=|x|$ at $x=0$ is a local minimum, and $f'(0)$ is undefined). Therefore, to find all potential locations of local extrema, we must examine all critical points.

Points where local extrema occur are always critical points. However, not all critical points correspond to local extrema.

Answer:

Local Maximum and Local Minimum Values (First and Second Derivative Tests)

Once we find the critical points of a function, we need tests to determine whether these points correspond to a local maximum, a local minimum, or neither. These tests use the information provided by the first and second derivatives about the slope and concavity of the function's graph.

First Derivative Test for Local Extrema

This test is based on the concept of monotonicity. By analyzing how the function changes from increasing to decreasing (or vice versa) around a critical point, we can classify the nature of that point.

Let $c$ be a critical point of a function $f$ (i.e., $f'(c)=0$ or $f'(c)$ is undefined, and $c$ is in the domain of $f$). Assume that $f$ is continuous at $c$.

Consider an interval $(p, q)$ containing $c$ where $c$ is the only critical point in that interval.

1. If $f'(x) > 0$ for all $x$ in $(p, c)$ and $f'(x) < 0$ for all $x$ in $(c, q)$, then $f$ has a local maximum at $c$.

   Interpretation: The function is increasing just before $c$ and decreasing just after $c$. This creates a peak at $c$.

2. If $f'(x) < 0$ for all $x$ in $(p, c)$ and $f'(x) > 0$ for all $x$ in $(c, q)$, then $f$ has a local minimum at $c$.

   Interpretation: The function is decreasing just before $c$ and increasing just after $c$. This creates a valley at $c$.

3. If $f'(x)$ has the same sign for all $x$ in $(p, c)$ and for all $x$ in $(c, q)$, then $f$ has no local extremum at $c$.

   Interpretation: If the function is increasing on both sides of $c$ or decreasing on both sides, $c$ is not a turning point that creates a local maximum or minimum. (This often occurs at inflection points with a horizontal tangent if $f'(c)=0$, like $f(x)=x^3$ at $x=0$).

To apply this test, you find the critical points, determine the intervals of monotonicity by testing the sign of $f'(x)$ in the intervals between critical points (and points of discontinuity), and then examine how the sign changes at each critical point.

Second Derivative Test for Local Extrema

The Second Derivative Test uses the concavity of the graph at a critical point where the tangent is horizontal ($f'(c)=0$). The sign of the second derivative $f''(c)$ tells us about the concavity at that point, which helps distinguish between a local maximum and a local minimum.

Suppose the second derivative $f''(x)$ is continuous on an open interval containing $c$, and $c$ is a critical point such that $f'(c) = 0$ (i.e., $c$ is a critical point with a horizontal tangent).

1. If $f''(c) > 0$, then $f$ has a local minimum at $c$.

   Interpretation: A positive second derivative means the graph is concave upward at $c$. If the tangent is horizontal at $c$ ($f'(c)=0$) and the curve is concave up ($f''(c)>0$), then $c$ must be the location of a local minimum (the bottom of a "bowl").

2. If $f''(c) < 0$, then $f$ has a local maximum at $c$.

   Interpretation: A negative second derivative means the graph is concave downward at $c$. If the tangent is horizontal at $c$ ($f'(c)=0$) and the curve is concave down ($f''(c)<0$), then $c$ must be the location of a local maximum (the top of a "dome").

3. If $f''(c) = 0$, the test is inconclusive. The critical point $c$ could be a local maximum, a local minimum, or an inflection point. In this situation, you must use the First Derivative Test to classify the critical point $c$.

Choosing Between the Tests:

• The First Derivative Test is more general as it applies to all critical points (where $f'(c)=0$ or $f'(c)$ is undefined). It also provides information about the function's increasing/decreasing intervals.
• The Second Derivative Test can be quicker if calculating $f''(x)$ and evaluating it at the critical points (where $f'(c)=0$) is easier than analyzing the sign of $f'(x)$ in intervals. However, it only applies to critical points with horizontal tangents and is inconclusive if $f''(c)=0$.

Both tests are valuable tools for classifying local extrema. The choice often depends on the specific function and critical points involved.

Answer:

Absolute (Global) Maximum and Minimum Values

While local extrema represent the highest or lowest points in a neighborhood, the absolute (or global) maximum and minimum values represent the overall highest and lowest function values over the entire domain of the function or a specified interval. These are the values we typically seek in optimization problems.

A function does not necessarily have an absolute maximum or minimum value over every interval or its entire domain (e.g., $f(x) = x^2$ on $(-\infty, \infty)$ has an absolute minimum but no absolute maximum; $f(x) = x$ on $(-\infty, \infty)$ has neither; $f(x)=1/x$ on $(0, \infty)$ has neither). However, there is a powerful theorem that guarantees the existence of absolute extrema under certain conditions.

Extreme Value Theorem (EVT)

The Extreme Value Theorem is a fundamental result in calculus that guarantees the existence of absolute extrema. It states:

If a function $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ is guaranteed to attain both an absolute maximum value and an absolute minimum value on that interval.

Furthermore, the theorem implies that these absolute extrema must occur at one of two types of points:

1. At a critical point of $f$ located within the open interval $(a, b)$, OR
2. At one of the endpoints of the interval ($x=a$ or $x=b$).

Note: The conditions of continuity and a closed interval are essential for the EVT. If the function is not continuous or the interval is open or unbounded, the function might not attain absolute extrema.

Procedure for Finding Absolute Extrema on a Closed Interval $[a, b]$ (Closed Interval Method)

The Extreme Value Theorem gives us a systematic procedure for finding the absolute maximum and minimum values of a continuous function on a closed interval $[a, b]$:

1. Verify Conditions: Ensure that the function $f$ is continuous on the closed interval $[a, b]$. If it is not, the EVT does not apply, and absolute extrema might not exist or may occur at points of discontinuity.
2. Find Critical Points within the Open Interval: Find all critical points of $f$ that lie *strictly within* the open interval $(a, b)$. These are the points $c$ such that $a < c < b$ and either $f'(c) = 0$ or $f'(c)$ is undefined.
3. Evaluate the function at the Critical Points: Calculate the value of $f(x)$ for each critical point found in Step 2.
4. Evaluate the function at the Endpoints: Calculate the value of $f(x)$ at the two endpoints of the interval: $f(a)$ and $f(b)$.
5. Compare all Values: Compare all the function values obtained in Step 3 (at critical points) and Step 4 (at endpoints). The largest value among these is the absolute maximum value of $f$ on $[a, b]$, and the smallest value among these is the absolute minimum value of $f$ on $[a, b]$.

This method is guaranteed to find the absolute extrema for any continuous function on a closed interval.

Answer:

Absolute Extrema on Open Intervals or Entire Domain

If the interval under consideration is open $(a, b)$, half-open $[a, b)$ or $(a, b]$, or is an infinite interval $(-\infty, b)$, $(a, \infty)$, or $(-\infty, \infty)$, the Extreme Value Theorem does not guarantee that an absolute maximum or minimum exists. A continuous function on such an interval may have no absolute extrema, one absolute extremum, or both.

To find absolute extrema on such intervals:

• Find Critical Points: Find all critical points of $f$ within the given interval.
• Analyze Function Behavior:
  • If there is only one critical point $c$ in the interval, and it corresponds to a local maximum (verified by the First or Second Derivative Test), then $f(c)$ is also the absolute maximum value of $f$ on that interval. Similarly, if the single critical point corresponds to a local minimum, it is the absolute minimum on the interval.
  • If there are multiple critical points, or if the single critical point does not yield the required extremum, you need to examine the behavior of the function as $x$ approaches the boundaries of the interval (which might be finite endpoints or $\pm \infty$) using limits.
  • Calculate $\lim\limits_{x \to a^+} f(x)$ (if interval starts at $a$), $\lim\limits_{x \to b^-} f(x)$ (if interval ends at $b$), $\lim\limits_{x \to \infty} f(x)$ (if interval extends to $\infty$), and $\lim\limits_{x \to -\infty} f(x)$ (if interval extends to $-\infty$).
  • Compare the function values at the critical points within the interval with the values of these limits. The largest value among the function values at critical points and any finite limit value is the absolute maximum (if a finite limit exists or if the function goes to $\infty$, there might not be an absolute max). Similarly for the absolute minimum. If limits go to $\pm \infty$, absolute extrema might not exist.

This process requires careful analysis of critical points and limits at the boundaries of the domain/interval.

Practical Problems on Maxima and Minima (Optimization Word Problems)

Many real-world problems can be formulated as optimization problems, where the goal is to find the maximum or minimum value of a quantity (like area, volume, cost, time, profit) subject to certain restrictions or conditions (constraints). Calculus, specifically the techniques for finding absolute extrema, provides a powerful framework for solving these problems.

General Strategy for Solving Optimization Problems

Solving optimization word problems involves translating the practical situation into a mathematical problem and then using calculus to find the desired extreme value. Follow these steps:

1. Understand the Problem:
   • Read the problem thoroughly. Identify precisely what quantity needs to be maximized or minimized (this is your objective).
   • Identify all the given information, conditions, or restrictions (these are your constraints).
2. Draw a Diagram (if applicable) and Introduce Variables:
   • If the problem involves geometry, draw a diagram and label it clearly.
   • Assign mathematical variables to the quantities that are changing or are unknown but relevant to the problem.
3. Formulate the Objective Function:
   • Write an equation that expresses the quantity you want to maximize or minimize (the objective quantity) in terms of the variables you defined in Step 2. This is your objective function.
4. Formulate the Constraint Equation(s):
   • Write down any constraints or relationships between the variables as equations. These are your constraint equations.
5. Express the Objective Function in Terms of a Single Variable:
   • If the objective function currently involves more than one variable, use the constraint equation(s) to eliminate all but one independent variable. Solve the constraint equation(s) for one variable and substitute that expression into the objective function.
   • You should end up with the objective function expressed as a function of a single independent variable (e.g., $A(x)$, $V(r)$, $C(q)$).
6. Determine the Feasible Domain of the Single Variable:
   • Based on the context of the problem, determine the set of possible values for the independent variable in your single-variable objective function. For instance, lengths, areas, volumes, or quantities of items must typically be non-negative ($x \ge 0$). Constraints might impose further restrictions, leading to a closed interval, an open interval, or an infinite interval.
7. Find the Absolute Extremum of the Objective Function over its Domain:
   • Use the calculus techniques for finding absolute extrema that are appropriate for the domain determined in Step 6.
   • Find the critical points of the objective function within the domain.
   • If the domain is a closed interval $[a, b]$ (which is often the case in basic problems), use the Closed Interval Method: evaluate the objective function at the critical points within $(a, b)$ and at the endpoints $a$ and $b$. The largest/smallest value is the absolute max/min.
   • If the domain is not a closed interval (e.g., open or infinite), use the First or Second Derivative Test to classify the critical points. If there is only one critical point in the domain and it corresponds to the desired local extremum (e.g., a local maximum when seeking absolute maximum), then it is also the absolute extremum. Otherwise, you may need to analyze limits as the variable approaches the boundaries of the domain.
8. Answer the Original Question:
   • Once you have found the value of the independent variable that optimizes the objective function, go back to the original problem. Calculate any other quantities or dimensions requested and state the final answer clearly, in the context of the problem, including appropriate units.

Answer:

Step 1: Understand the Problem.

Objective: Maximize the area of the rectangular garden.

Constraint: A total of 100 meters of fencing is available for three sides of the garden (one side is along a river).

Step 2: Diagram and Variables.

Draw a rectangle representing the garden. Let the side parallel to the river have length $y$ meters, and the two sides perpendicular to the river each have length $x$ meters.

Step 3: Objective Function.

The area of the rectangular garden is $A = \text{length} \times \text{width}$. In terms of our variables, this is:

"$A = xy$"

This is the function we want to maximize.

Step 4: Constraint Equation(s).

The total length of the fencing is 100 meters, and it's used for the two sides of length $x$ and the one side of length $y$.

"$2x + y = 100$"

[Constraint equation]

Step 5: Single Variable Objective Function.

The objective function $A = xy$ has two variables, $x$ and $y$. Use the constraint equation to eliminate one variable. It's usually easier to solve for $y$ in the constraint equation:

"$y = 100 - 2x$"

[Solve constraint for $y$]

Substitute this expression for $y$ into the area formula:

"$A(x) = x(100 - 2x)$"

[Substitute into area formula]

Expand the expression:

"$A(x) = 100x - 2x^2$"

Now, the area is expressed as a function of a single variable $x$.

Step 6: Determine the Feasible Domain.

The dimensions of the garden must be non-negative. So, $x \ge 0$ and $y \ge 0$.

From $x \ge 0$, we have a lower bound for the domain of $x$.

From $y = 100 - 2x \ge 0$, we get $100 \ge 2x$, which implies $50 \ge x$. So, $x \le 50$.

Considering $x$ must also be greater than 0 (a garden with zero width has no area), the feasible domain for $x$ is the closed interval $[0, 50]$.

Domain of $A(x)$: $[0, 50]$.

Step 7: Find the Absolute Maximum.

We need to find the absolute maximum value of $A(x) = 100x - 2x^2$ on the closed interval $[0, 50]$. Since $A(x)$ is a polynomial, it is continuous on this closed interval. Thus, the Extreme Value Theorem applies, and the absolute maximum will occur at a critical point in $(0, 50)$ or at the endpoints $x=0$ or $x=50$.

First, find the derivative $A'(x)$:

"$A'(x) = \frac{d}{dx}(100x - 2x^2) = 100 - 4x$"

Find critical points in $(0, 50)$ by setting $A'(x) = 0$:

"$100 - 4x = 0$"

"$4x = 100 \implies x = 25$"

The critical point is $x = 25$. This point lies within the open interval $(0, 50)$.

Now, evaluate the area function $A(x)$ at the critical point(s) in the interval and at the endpoints of the interval.

• At critical point $x = 25$:

  $A(25) = 100(25) - 2(25)^2 = 2500 - 2(625) = 2500 - 1250 = 1250$.

• At endpoint $x = 0$:

  $A(0) = 100(0) - 2(0)^2 = 0 - 0 = 0$.

• At endpoint $x = 50$:

  $A(50) = 100(50) - 2(50)^2 = 5000 - 2(2500) = 5000 - 5000 = 0$.

Compare the values: $A(25) = 1250$, $A(0) = 0$, $A(50) = 0$. The largest value is 1250.

The maximum area is 1250 m², which occurs when $x = 25$ meters.

Step 8: Answer the Question.

The question asks for the dimensions that maximize the area. We found that the maximum area occurs when $x = 25$ meters.

Find the corresponding dimension $y$ using the constraint equation $y = 100 - 2x$:

"$y = 100 - 2(25) = 100 - 50 = 50$"

So, $y = 50$ meters.

The dimensions of the garden that will maximize the enclosed area are 25 meters for the sides perpendicular to the river and 50 meters for the side parallel to the river.

Maxima and Minima (Applied Maths)

From an applied mathematics viewpoint, finding maxima and minima using derivatives is a cornerstone technique for optimization.

Optimization is the process of finding the best possible solution to a problem, where "best" means maximizing or minimizing some quantity (like profit, efficiency, cost, strength, time). Derivatives provide a systematic analytical method to find these optimal values, complementing graphical or numerical approaches.

Key Concepts and Interpretation in Applied Problems

Applying calculus to solve real-world optimization problems involves several key steps and interpretations:

• Mathematical Modeling: The first critical step is to translate the practical problem into a mathematical model. This involves identifying the quantities involved, assigning variables, and formulating mathematical relationships between them.
• Objective Function: This is the function that represents the quantity to be maximized or minimized. It must be clearly defined in terms of the variables chosen. Examples include cost functions, revenue functions, profit functions, area formulas, volume formulas, distance formulas, time functions, etc. The goal is to find the input value(s) that give the extreme output value of this function.
• Constraints: Real-world problems always have limitations or conditions that must be satisfied. These constraints are formulated as equations or inequalities relating the variables. Constraints are essential because they limit the feasible values of the variables and often allow the objective function to be expressed in terms of a single variable.
• Domain of Feasibility: Based on the context of the problem and the variables chosen, determine the realistic range of values that the independent variable in the single-variable objective function can take. Physical quantities like lengths, weights, times, quantities of items, etc., are typically non-negative. Constraints can further restrict this domain. Defining this domain correctly is crucial for finding the absolute extremum.
• Calculus Tools for Finding Extrema: The techniques developed for finding local and absolute extrema become the tools for optimization:
  • Finding Critical Points: Calculating the first derivative of the objective function ($f'(x)$) and finding where it is zero or undefined helps locate potential points of maximum or minimum. These represent points where the rate of change of the objective quantity is zero or where the model might not be smooth.
  • Classifying Critical Points: Using the First Derivative Test (sign change of $f'(x)$) or the Second Derivative Test (sign of $f''(x)$ at points where $f'(x)=0$) helps determine if a critical point is a local maximum, local minimum, or neither.
  • Finding Absolute Extrema: Depending on the nature of the feasible domain (closed interval, open interval, infinite), applying the appropriate method (Closed Interval Method for $[a,b]$ or limit analysis/single critical point test for other intervals) ensures finding the overall best possible value.
• Interpretation of the Solution: The final mathematical answer (the value of the independent variable that optimizes the objective function) must be translated back into the context of the original problem. This involves stating the dimensions, quantities, times, etc., that achieve the maximum or minimum, and stating the resulting maximum or minimum value of the objective quantity itself. Including appropriate units is vital for a meaningful applied answer.

The process of setting up the model, applying calculus techniques, and interpreting the results allows applied mathematicians to solve a vast range of optimization problems, providing quantitative answers to questions about maximizing efficiency, minimizing costs, and designing optimal systems.