5 Optimization and Decision Models

A progressive guide to formulating, solving, and interpreting optimization models with one or more variables, constraints, boundaries, and discrete decisions.

The Structure of an Optimization Model

Optimization is the systematic search for the best feasible choice. A decision model translates a real situation into mathematics so that a quantity such as cost, profit, production time, area, or risk can be maximized or minimized.

A complete model contains four connected parts:

Decision variables: quantities under the decision maker’s control.
: the quantity to maximize or minimize.
Constraints: equations or inequalities that restrict allowable choices.
: all choices that satisfy every .

For example, if a manufacturer produces quantities $x$ and $y$ of two products, with unit profits of $\$40$ and $\$30$ , the profit objective is

P(x,y)=40x+30y.

If machine time imposes the limitation $2x+y\leq 100$ , and production quantities cannot be negative, the model is

\text{Maximize } P(x,y)=40x+30y

subject to

2x+y\leq 100,\qquad x\geq 0,\qquad y\geq 0.

A choice is feasible when it satisfies all restrictions. An is the feasible choice with the best objective value under the stated model.

Takeaway: Optimization is not just differentiation or computation; it begins by identifying the decision, the goal, and every restriction that defines the .

From an Applied Situation to a Model

A reliable formulation process connects the real situation to a mathematical model:

Identify the decision. Determine what can be changed.
Define variables and units. State precisely what each variable represents.
Define the . Decide whether the goal is maximization or minimization.
Translate limitations into constraints. Include physical, financial, technical, and nonnegativity restrictions.
Determine the domain. Exclude mathematically possible values that are impossible in the application.
Solve and compare candidates. Evaluate the objective at relevant critical points, boundary points, endpoints, and corners.
Interpret the result. Report the decision, objective value, units, and practical meaning.

For instance, a model may allow a variable to be any real number mathematically, but the application may require it to be nonnegative, bounded, or an integer. These requirements belong in the model before solving.

A derivative can optimize only the function that has been constructed. It cannot repair an incorrectly defined or restore a that was omitted.

Takeaway: Careful formulation is part of the mathematics. A correct solution to an incorrect model is not a correct solution to the real decision problem.

One-Variable Optimization

When the constraints allow one variable to be eliminated, express the objective as a one-variable function on a physically meaningful interval. For a differentiable function, interior extrema can occur where

f'(x)=0.

A is a value in the domain where $f'(x)=0$ or where $f'(x)$ does not exist. On a closed interval $[a,b]$ , an absolute maximum or minimum can occur at an interior , the left endpoint $a$ , or the right endpoint $b$ .

The standard procedure is:

Determine the valid closed interval.
Find all critical numbers in the interior.
Evaluate the objective at every and both endpoints.
Compare the values and interpret the best candidate.

Example: Garden Along a Wall

Suppose three sides of a rectangular garden require fencing, while the fourth side borders a wall. Let $x$ be the width perpendicular to the wall and $y$ the length along the wall. With $100$ meters of fencing,

2x+y=100, \qquad y=100-2x.

The area becomes

A(x)=xy=x(100-2x)=100x-2x^2.

Because lengths cannot be negative,

0\leq x\leq 50.

Differentiate:

A'(x)=100-4x.

The interior satisfies

100-4x=0, \qquad x=25.

Compare the candidates:

A(0)=0, \qquad A(25)=1250, \qquad A(50)=0.

Therefore, the maximum area is $1250\text{ m}^2$ , achieved when

x=25\text{ m}, \qquad y=50\text{ m}.

The complete conclusion includes both the dimensions and the maximum area, not merely the value $x=25$ .

Takeaway: On a closed interval, never stop after finding a derivative equal to zero; compare every interior with all endpoints.

Interior Critical Points and Local Behavior

For a function of two variables, an interior critical point occurs where both first partial derivatives vanish or where one of them does not exist:

f_x(x,y)=0, \qquad f_y(x,y)=0.

For a twice-differentiable function, define

D=f_{xx}f_{yy}-(f_{xy})^2.

At a critical point:

If $D>0$ and $f_{xx}>0$ , the point is a local minimum.
If $D>0$ and $f_{xx}<0$ , the point is a local maximum.
If $D<0$ , the point is a saddle point.
If $D=0$ , the test is inconclusive.

A local optimum is best only in a neighborhood. It is not automatically the best feasible point across the entire region. Global conclusions require comparison with the boundary and any other relevant candidates.

Takeaway: The multivariable second-derivative test classifies local behavior, but constrained optimization still requires a global search over the feasible set.

Boundary Analysis

For a constrained problem, an optimum may occur where one or more constraints hold with equality. This boundary can contain the global maximum or minimum even when no interior critical point exists.

For a closed, bounded region:

Find all critical points in the interior.
Analyze each boundary segment or boundary curve.
Find critical points along each boundary component.
Check endpoints and corner points.
Evaluate the objective at every candidate and compare the values.

Consider

f(x,y)=x+y

on the triangular region

x\geq 0, \qquad y\geq 0, \qquad x+y\leq 10.

There is no interior critical point because

f_x=1, \qquad f_y=1,

so the two partial derivatives cannot both equal zero. On the boundary $x+y=10$ , however,

f(x,y)=10.

Thus, every feasible point on that boundary has the same maximum value. The minimum occurs at the corner $(0,0)$ , where

f(0,0)=0.

Takeaway: Ignoring the boundary can miss both global extrema. Interior analysis is only one part of constrained optimization.

Equality Constraints and Lagrange Multipliers

When the objective $f(x,y)$ is optimized subject to an equality $g(x,y)=c$ , the method of Lagrange multipliers finds candidate points by solving

\nabla f=\lambda\nabla g,

together with

g(x,y)=c.

Equivalently,

f_x=\lambda g_x, \qquad f_y=\lambda g_y, \qquad g(x,y)=c.

A is the parameter $\lambda$ in this system. Geometrically, the objective’s level curve and the curve are tangent at a constrained optimum, so their gradients are parallel. The method generates candidates, but those candidates must still be evaluated and compared. Inequality constraints, endpoints, and other boundary components may require separate analysis.

Example: Maximum Rectangle Area with Fixed Perimeter

For a rectangle with side lengths $x$ and $y$ , the area is

A=xy,

and a fixed perimeter $P$ gives the

2x+2y=P.

Applying the equations produces

x=y=\frac{P}{4}.

Therefore, among rectangles with fixed perimeter, the square has the greatest area.

Takeaway: Lagrange multipliers identify constrained candidates through parallel gradients; they do not eliminate the need to verify feasibility and compare candidates.

Model Types and Interpreting the Result

Optimization models also differ according to the forms of their functions and the allowed values of their variables.

A linear program has a linear objective and linear constraints.
A nonlinear program has a nonlinear objective, nonlinear constraints, or both.
An requires some or all decision variables to be integers.
A restricts selected variables to values such as $0$ or $1$ , often representing yes-or-no decisions.

Variable type is part of feasibility. For example, a computed value of $3.6$ buses is not directly feasible when the decision variable counts whole buses. Rounding can violate constraints or produce a nonoptimal result, so integer restrictions should be included from the beginning.

A model’s optimum is conditional on its objective, constraints, data, and assumptions. A complete interpretation should state:

The values of the decision variables.
The resulting maximum or minimum and its units.
Which constraints are active, meaning they hold with equality.
Which assumptions limit the conclusion, such as continuous production, constant prices, fixed resources, or an assumed functional relationship.

If an input changes, the model should be reevaluated. A mathematically optimal decision may also require practical checks involving uncertainty, safety, fairness, capacity, or implementation cost.

Takeaway: A strong optimization conclusion reports what to do, how good the result is, which limitations bind, and under what assumptions the recommendation is valid.