Optimization from a Model

The derivative is not the first step

An optimization problem begins with a quantity to improve and constraints on what is possible. Name the variables, attach units and draw a diagram if geometry is involved. Use the constraints to reduce the objective to one variable, then state the feasible domain.

Only after this modeling work should you differentiate. Solve for interior critical points, examine endpoints or limiting boundary behavior, and justify why the selected candidate is the desired global optimum. A numerical root of a derivative is not itself that justification.

Visual guide

VISUAL GUIDEA constraint turns geometry into one variable

Area x(6 − x)

A rectangle with perimeter 12 has sides x and 6 − x, so area A = x(6 − x) for 0 ≤ x ≤ 6. Its highest point is the square x = 3, A = 9. Endpoints represent collapsed rectangles with zero area.

Worked example: a fence with one free side

A rectangular enclosure stands against a straight wall. There are $40$ meters of fencing for the other three sides. Let $x$ be the perpendicular width and $y$ the side parallel to the wall. The constraint is $2x+y=40$ , so $y=40-2x$ .

Area is $A(x)=x(40-2x)$ with $0\le x\le20$ if degenerate rectangles are admitted as boundary cases. Then $A'=40-4x$ , giving $x=10$ . The corresponding $y$ is $20$ and the area is $200$ m². The second derivative is $-4$ , and the endpoint areas are zero. The concave quadratic therefore has its global maximum at this point.

Using $2x+2y=40$ would solve a different problem. The physical constraint must be correct before the calculus can be meaningful.

Worked example: minimum material for an open box

A square-base open box must hold $32$ volume units. With base side $x>0$ , its height is $h=32/x^2$ . Material area is

S(x)=x^2+4xh=x^2+\frac{128}{x}.

Differentiate: $S'=2x-128/x^2$ . Setting this to zero gives $2x^3=128$ , hence $x=4$ and $h=2$ . Also $S''=2+256/x^3>0$ for all feasible $x$ . Since $S$ grows without bound as $x\to0^+$ or $x\to\infty$ , this is the global minimum, with area $48$ .

The domain is open and unbounded, so the closed-interval recipe alone is not enough. Boundary limits supply the missing global check.

Sanity checks and limitations

Return to the original variables and verify the constraint. Check signs, dimensions and whether the solution is physically admissible. Real manufacturing may impose integer dimensions or minimum wall thickness; those constraints can alter the mathematical optimum. State the model rather than silently solving a more realistic problem than the information permits.

Practice

A rectangle has perimeter $24$ . Maximize its area.
Minimize $x+9/x$ for $x>0$ .
Why must endpoints be checked even after finding an interior stationary point?

Show worked solutions

With sides $x$ and $12-x$ , area $12x-x^2$ is maximized at $x=6$ . The square has area $36$ .
The derivative is $1-9/x^2$ , negative below $3$ and positive above. The minimum is $6$ at $x=3$ .
A stationary point can be a minimum while the maximum occurs at a boundary, as with $x^2$ on $[-1,2]$ .