The Central Idea
An optimization problem combines two ingredients:
- an objective: the quantity to maximize or minimize;
- a constraint: a condition limiting the possible choices.
The general structure is
\[ \boxed{ \text{constraint} \longrightarrow \text{one-variable objective function} \longrightarrow \text{critical points and endpoints} \longrightarrow \text{best feasible value} }. \]What Does Optimization Mean?
To optimize a quantity means to find a largest or smallest feasible value according to the problem's conditions.
Typical goals include:
- maximize area;
- maximize volume;
- maximize revenue or profit;
- minimize cost;
- minimize distance or travel time;
- minimize material used;
- maximize output subject to limited resources.
The Objective Function
The quantity being optimized is represented by an objective function.
For example, if a rectangle has length \(L\) and width \(W\), its area is
\[ A=LW. \]If the goal is to maximize area, then
\[ \boxed{A} \]is the objective quantity.
The Constraint
A constraint is a condition that limits which values are allowed.
If a rectangle has perimeter \(40\), then
\[ 2L+2W=40. \]This equation links the two dimensions and prevents them from being chosen independently.
Why We Often Reduce to One Variable
Introductory derivative methods are easiest to apply when the objective is written as a function of one variable.
From
\[ 2L+2W=40, \]divide by \(2\):
\[ L+W=20. \]Solve for \(W\):
\[ W=20-L. \]Substitute into the area formula:
\[ A(L)=L(20-L). \]The Feasible Domain
The algebraic formula is not the entire problem. Physical or logical conditions restrict the allowable inputs.
For the rectangle,
\[ L\ge0 \]and
\[ W=20-L\ge0. \]Therefore,
\[ \boxed{0\le L\le20}. \]This interval is the feasible domain for the model.
Optimization Is Extrema Analysis with Context
Once the objective function and feasible domain are known, optimization becomes an extrema problem.
We ask:
- Where is the derivative zero?
- Where is the derivative undefined?
- Which candidates lie in the feasible domain?
- Do endpoints need to be checked?
- Which candidate gives the largest or smallest objective value?
A Basic Optimization Workflow
- Identify the quantity to optimize.
- Write an objective formula.
- Write the constraint.
- Use the constraint to rewrite the objective in one variable.
- Determine the feasible domain.
- Differentiate the objective function.
- Find critical numbers in the feasible domain.
- Check endpoints when appropriate.
- Compare objective values.
- Interpret the result in the original context.
Example: Maximum Area with Fixed Perimeter
A rectangle has perimeter \(40\). Find the dimensions that maximize its area.
Objective:
\[ A=LW. \]Constraint:
\[ 2L+2W=40. \]Solve the constraint:
\[ W=20-L. \]Build the One-Variable Objective
Substitute
\[ W=20-L \]into
\[ A=LW. \]Then
\[ A(L)=L(20-L). \]Expand:
\[ \boxed{ A(L)=20L-L^2 }. \]The feasible interval is
\[ 0\le L\le20. \]Differentiate the Area Function
From
\[ A(L)=20L-L^2, \]differentiate:
\[ A'(L)=20-2L. \]Set equal to zero:
\[ 20-2L=0. \]Therefore,
\[ \boxed{L=10}. \]Recover the Other Dimension
Since
\[ W=20-L, \]substitute \(L=10\):
\[ W=20-10=10. \]Thus, the candidate rectangle is
\[ \boxed{10\times10}. \]Verify the Maximum
The second derivative is
\[ A''(L)=-2. \]Since
\[ A''(10)<0, \]the area function is concave down, so the critical point gives a local maximum.
On the closed feasible interval, we may also compare endpoints:
\[ A(0)=0, \qquad A(10)=100, \qquad A(20)=0. \]Therefore, the maximum area is
\[ \boxed{100} \]square units.
Interpret the Rectangle Result
Among all rectangles with perimeter \(40\), the largest area occurs when the rectangle is a square.
\[ \boxed{ L=W=10 }. \]An optimization solution should end with this contextual interpretation, not merely the critical number \(L=10\).
Critical Number versus Optimal Value
In the rectangle example,
\[ L=10 \]is the critical input.
The optimized quantity is
\[ A(10)=100. \]The optimal dimensions are
\[ 10\times10. \]These are three related but different outputs of the problem.
Maximum and Minimum Problems
The derivative process is similar whether the objective is being maximized or minimized.
What changes is the final comparison.
- For a maximum problem, choose the greatest feasible objective value.
- For a minimum problem, choose the least feasible objective value.
Example: Minimum of a Quadratic Cost Function
Suppose a simplified cost model is
\[ C(x)=x^2-12x+50, \]where \(x\) lies in the feasible interval
\[ 0\le x\le10. \]Differentiate:
\[ C'(x)=2x-12. \]Set equal to zero:
\[ 2x-12=0. \]Thus,
\[ \boxed{x=6}. \]Check the Minimum Cost
Evaluate the objective at the critical number and endpoints:
\[ C(0)=50, \] \[ C(6)=36-72+50=14, \] \[ C(10)=100-120+50=30. \]Therefore, the minimum cost is
\[ \boxed{14} \]at
\[ \boxed{x=6}. \]Why Endpoints Must Be Checked
A critical point can give a local extremum without giving the best value on a restricted interval.
On a closed interval
\[ [a,b], \]the absolute maximum or minimum may occur at:
- an interior critical number;
- the left endpoint \(a\);
- the right endpoint \(b\).
Example: The Endpoint Wins
Suppose
\[ f(x)=x^3-3x \]on
\[ [0,3]. \]Differentiate:
\[ f'(x)=3x^2-3. \]The interior critical number is
\[ x=1. \]Evaluate:
\[ f(0)=0, \qquad f(1)=-2, \qquad f(3)=18. \]The absolute maximum occurs at the endpoint
\[ \boxed{x=3}. \]Objective Functions May Begin with Several Variables
Many practical formulas contain more than one variable.
For example, the volume of a rectangular box is
\[ V=LWH. \]If the problem provides relationships among \(L\), \(W\), and \(H\), those constraints can sometimes be used to rewrite \(V\) as a function of one variable.
The central modeling task is often the reduction from several variables to one.
Example: Rectangle Against a Wall
Suppose \(100\) meters of fencing are available for three sides of a rectangular enclosure because the fourth side lies along a wall.
Let \(x\) be the width perpendicular to the wall and \(y\) the length along the wall.
The fencing constraint is
\[ 2x+y=100. \]The area is
\[ A=xy. \]Reduce the Enclosure Problem to One Variable
Solve the constraint:
\[ y=100-2x. \]Substitute into the area:
\[ A(x)=x(100-2x). \]Therefore,
\[ \boxed{ A(x)=100x-2x^2 }. \]Since both dimensions must be nonnegative,
\[ 0\le x\le50. \]Optimize the Enclosure Area
Differentiate:
\[ A'(x)=100-4x. \]Set equal to zero:
\[ 100-4x=0. \]Thus,
\[ x=25. \]Recover \(y\):
\[ y=100-2(25)=50. \]Therefore, the maximum-area dimensions are
\[ \boxed{x=25,\qquad y=50}. \]Maximum Area of the Enclosure
Evaluate:
\[ A(25)=25(50)=1250. \]Therefore, the maximum area is
\[ \boxed{1250\text{ m}^2}. \]Notice that the optimal rectangle is not a square because the fencing constraint treats the sides differently.
Optimization Depends on the Constraint
Two problems can optimize the same quantity but produce different answers because their constraints differ.
For a four-sided rectangle with fixed perimeter, the optimal shape is a square.
For a three-sided enclosure against a wall, the optimal dimensions satisfy
\[ y=2x. \]The objective formula alone does not determine the optimum; the constraint is part of the mathematics.
Units Matter in Optimization
The optimized quantity carries units appropriate to the objective.
- length: meters, feet, kilometers;
- area: square units;
- volume: cubic units;
- cost: currency;
- rate: output units per input unit.
If the result is an area, reporting only
\[ 1250 \]without square units leaves the interpretation incomplete.
Feasibility Comes Before Optimality
A mathematically valid critical number may be unusable in the original problem.
Suppose a length calculation produces
\[ x=-4. \]If \(x\) represents a physical length, the value is infeasible and must be rejected.
Optimization always occurs within the feasible set.
Example: Rejecting an Infeasible Candidate
Suppose an objective function has critical numbers
\[ x=-2 \]and
\[ x=5, \]but the model requires
\[ 0\le x\le8. \]Then only
\[ \boxed{x=5} \]is a feasible critical number.
The Second Derivative Can Confirm an Optimum
If
\[ f'(c)=0, \]then:
- \(f''(c)>0\) supports a local minimum;
- \(f''(c)<0\) supports a local maximum;
- \(f''(c)=0\) makes the second derivative test inconclusive.
In closed-interval optimization, endpoint comparison is still required when absolute extrema are sought.
Factoring Can Simplify Optimization
Suppose
\[ P(x)=-x^3+12x^2. \]Differentiate:
\[ P'(x)=-3x^2+24x. \]Factor:
\[ P'(x)=-3x(x-8). \]The critical numbers are
\[ x=0,\ 8. \]Factored derivatives make candidate values easier to identify.
Optimization with a Revenue Function
Suppose a simplified revenue model is
\[ R(x)=-2x^2+120x, \]for
\[ 0\le x\le60. \]Differentiate:
\[ R'(x)=-4x+120. \]Set equal to zero:
\[ -4x+120=0. \]Thus,
\[ \boxed{x=30}. \]Maximum Revenue
Evaluate:
\[ R(30) = -2(30)^2+120(30). \] \[ = -1800+3600 = 1800. \]Since the quadratic opens downward, this is the maximum.
\[ \boxed{ R_{\max}=1800 }. \]The units depend on the business model used to define \(R\).
Profit Is Not the Same as Revenue
Revenue measures incoming money from sales.
Profit accounts for costs:
\[ \boxed{ P(x)=R(x)-C(x) }. \]A quantity that maximizes revenue does not necessarily maximize profit.
The objective function must match the question actually being asked.
Optimization Can Be Geometric or Applied
The same mathematical workflow appears in many settings.
- geometry: maximize area or volume;
- economics: maximize revenue or profit;
- engineering: minimize material or energy;
- transportation: minimize distance or time;
- design: choose dimensions under space constraints;
- science: maximize or minimize a modeled response.
The context changes, but the derivative-based structure remains similar.
Optimization Is a Modeling Process
Differentiation is only one part of optimization.
A correct derivative cannot rescue an incorrect model.
The hardest steps are often:
- choosing variables;
- identifying the objective;
- translating the constraint;
- determining the feasible domain;
- interpreting the mathematical result correctly.
Choose Variables Clearly
Before building equations, define what each variable represents.
For example:
\[ x=\text{width of the enclosure in meters} \]and
\[ y=\text{length of the enclosure in meters}. \]Clear variable definitions reduce ambiguity and help preserve units.
Translate the Constraint Carefully
In the three-sided fencing problem, only two widths and one length require fencing.
Therefore,
\[ 2x+y=100, \]not
\[ 2x+2y=100. \]A small modeling error changes the entire optimization problem.
Common Error: Optimizing the Constraint Instead of the Objective
The constraint limits the choices but is not necessarily the quantity being optimized.
For example,
\[ 2x+y=100 \]is the fencing constraint.
The quantity to maximize is
\[ A=xy. \]Differentiate the objective after using the constraint to rewrite it.
Common Error: Differentiating Before Reducing Variables
If the objective is
\[ A=xy \]but both \(x\) and \(y\) are still independent symbols, elementary one-variable differentiation does not yet solve the problem.
Use the constraint to express one variable in terms of the other first.
Common Error: Ignoring the Feasible Domain
An algebraic objective may be defined for values that make no sense in the original context.
Negative lengths, negative production quantities, or dimensions exceeding a resource limit must be rejected.
Always determine the feasible interval before choosing an optimum.
Common Error: Assuming Every Critical Number Is Optimal
A critical number is only a candidate.
It may correspond to:
- a local maximum;
- a local minimum;
- neither;
- an inferior value compared with an endpoint.
Classification and comparison are necessary.
Common Error: Forgetting to Recover the Original Variables
Suppose optimization gives
\[ x=25 \]and the constraint is
\[ y=100-2x. \]The solution is incomplete until
\[ y=50 \]is also found if the problem asks for dimensions.
Common Error: Reporting the Critical Input Instead of the Optimized Quantity
If
\[ A(25)=1250, \]then \(25\) is an input dimension, while \(1250\) is the area.
If the question asks for maximum area, the answer is
\[ \boxed{1250\text{ square units}}, \]not merely \(25\).
Common Error: Ignoring Units
Optimization often combines quantities with different dimensions.
A length answer should use linear units, an area answer square units, and a volume answer cubic units.
Units are part of the mathematical interpretation, not decoration.
Common Error: Assuming a Maximum Exists
Not every optimization problem has an attained maximum or minimum.
On an open or unbounded domain, a function may increase indefinitely or approach a limiting value without reaching it.
Always consider the domain and the actual behavior of the objective function.
A Reliable Procedure for Geometric Optimization
- Draw or interpret the geometric configuration.
- Define the dimensions with variables.
- Write the objective formula, such as area or volume.
- Write the geometric or resource constraint.
- Solve the constraint for one variable.
- Substitute into the objective.
- Determine the feasible domain.
- Differentiate and find critical numbers.
- Check endpoints or classify candidates.
- Recover all requested dimensions and state the optimized value with units.
A Reliable Procedure for Applied Optimization
- Identify exactly what quantity the problem asks to maximize or minimize.
- Define variables and units.
- Build the objective function.
- Translate every constraint.
- Reduce the model to one independent variable when possible.
- State the feasible interval.
- Differentiate.
- Find and filter critical numbers.
- Compare all relevant candidates.
- Interpret the final answer in the original context.
Example: Full Optimization Workflow
A rectangle has perimeter \(60\). Find the maximum possible area.
Let the dimensions be \(x\) and \(y\).
Constraint:
\[ 2x+2y=60. \]Therefore,
\[ x+y=30 \]and
\[ y=30-x. \]Objective:
\[ A=xy. \]Substitute:
\[ A(x)=x(30-x)=30x-x^2. \]Complete the Full Optimization Workflow
The feasible interval is
\[ 0\le x\le30. \]Differentiate:
\[ A'(x)=30-2x. \]Set equal to zero:
\[ 30-2x=0. \]Thus,
\[ x=15. \]Then
\[ y=30-15=15. \]The maximum area is
\[ A(15)=225. \]Therefore,
\[ \boxed{ \text{maximum area}=225\text{ square units} } \]with dimensions
\[ \boxed{15\times15}. \]Practice 1: Maximum Area
A rectangle has perimeter \(24\).
Find the dimensions that maximize area and find the maximum area.
Practice 2: Three-Sided Enclosure
A farmer has \(60\) meters of fencing for three sides of a rectangle, with the fourth side along a wall.
Find the dimensions that maximize area.
Practice 3: Minimum Quadratic Cost
Let
\[ C(x)=x^2-10x+40, \]where
\[ 0\le x\le8. \]Find the minimum value of \(C\).
Practice 4: Maximum Revenue
Let
\[ R(x)=-x^2+40x \]for
\[ 0\le x\le40. \]Find the maximum revenue value.
Practice 5: Endpoint Comparison
Let
\[ f(x)=x^3-6x^2+9x \]on
\[ [0,5]. \]Find the absolute maximum and absolute minimum values.
Practice 6: Feasible Candidate
An optimization model has critical numbers
\[ -3,\ 4,\ 11 \]and feasible domain
\[ 0\le x\le8. \]Which critical number remains feasible?
Practice 7: Recover a Second Variable
A constraint is
\[ 2x+y=80. \]If optimization gives
\[ x=20, \]find \(y\).
Practice 8: Identify the Objective
A rectangular garden must use \(50\) meters of fencing and should have the greatest possible area.
Which formula is the objective, and which formula is the constraint?
Practice 9: Interpret the Units
An optimization calculation gives a maximum rectangular area of
\[ 360. \]If all lengths were measured in meters, what units should be attached to the answer?
Practice 10: Identify the Error
A student finds a critical number \(x=12\) in a model whose feasible interval is
\[ 0\le x\le10 \]and reports \(x=12\) as the optimum.
Explain the error.
Practice Answers
Practice 1
Let the sides be \(x\) and \(y\).
\[ 2x+2y=24 \] \[ x+y=12 \] \[ y=12-x. \]Area:
\[ A(x)=x(12-x)=12x-x^2. \]Differentiate:
\[ A'(x)=12-2x. \]Set equal to zero:
\[ x=6. \]Then \(y=6\). Therefore,
\[ \boxed{6\times6} \]gives maximum area
\[ \boxed{36\text{ square units}}. \]Practice 2
Let \(x\) be each perpendicular side and \(y\) the side parallel to the wall.
\[ 2x+y=60. \] \[ y=60-2x. \]Area:
\[ A(x)=x(60-2x)=60x-2x^2. \]Differentiate:
\[ A'(x)=60-4x. \]Set equal to zero:
\[ x=15. \]Then
\[ y=60-30=30. \]Therefore, the maximizing dimensions are
\[ \boxed{15\text{ m by }30\text{ m}}. \]Practice 3
\[ C'(x)=2x-10. \]Critical number:
\[ x=5. \]Evaluate:
\[ C(0)=40, \qquad C(5)=25-50+40=15, \qquad C(8)=64-80+40=24. \]Therefore, the minimum value is
\[ \boxed{15}. \]Practice 4
\[ R'(x)=-2x+40. \]Set equal to zero:
\[ x=20. \]Then
\[ R(20)=-400+800=400. \]Therefore,
\[ \boxed{R_{\max}=400}. \]Practice 5
\[ f'(x)=3x^2-12x+9. \] \[ =3(x-1)(x-3). \]Interior critical numbers:
\[ x=1,\ 3. \]Evaluate:
\[ f(0)=0, \] \[ f(1)=1-6+9=4, \] \[ f(3)=27-54+27=0, \] \[ f(5)=125-150+45=20. \]Therefore, the absolute minimum value is
\[ \boxed{0} \]and the absolute maximum value is
\[ \boxed{20}. \]Practice 6
Only
\[ \boxed{x=4} \]lies in the feasible interval \(0\le x\le8\).
Practice 7
\[ 2(20)+y=80. \] \[ 40+y=80. \] \[ \boxed{y=40}. \]Practice 8
The objective is the area:
\[ \boxed{A=LW}. \]The constraint is the perimeter:
\[ \boxed{2L+2W=50}. \]Practice 9
Area uses square units. Therefore,
\[ \boxed{360\text{ m}^2}. \]Practice 10
The critical number
\[ x=12 \]lies outside the feasible interval
\[ 0\le x\le10. \]Therefore, it cannot be the optimum for the stated problem. Only feasible critical numbers and relevant endpoints may be considered.
What to Remember
Optimization combines calculus with modeling.
\[ \boxed{ \text{objective} + \text{constraint} + \text{feasible domain} + \text{extrema analysis} = \text{optimization} }. \]- identify exactly what is being maximized or minimized;
- write the objective function;
- translate the constraint accurately;
- reduce the objective to one variable when possible;
- determine the feasible domain before selecting a solution;
- find critical numbers using the derivative;
- check endpoints for closed-interval absolute extrema;
- reject infeasible candidates;
- recover any eliminated variables;
- report the optimized quantity with appropriate units and context.
Where We Go Next
Derivatives describe how quantities change and help locate their largest and smallest values. The next major idea reverses perspective: instead of asking for the rate at one point, we begin combining many small contributions over an interval to measure total accumulation.
The next lesson, Area and Accumulated Change, introduces area under a curve as a model of accumulation and begins the transition from differential ideas toward integral ideas.