Optimization as a Modeling Problem

An optimization problem does not begin with a derivative. It begins by deciding what is being optimized, which quantities may change, and what restrictions connect them. Different assumptions create different mathematics even when the objects look similar.

A rectangle beside a wall

A rectangular enclosure needs fencing on three sides because an existing wall supplies the fourth. Let $x$ be each perpendicular side and $y$ the parallel fenced side, in metres. With $60$ metres of fencing, $2x+y=60$. The objective is area $A=xy$. Substituting the constraint gives

$$A(x)=x(60-2x),\qquad 0\leq x\leq30.$$

The closed interval includes degenerate rectangles so that boundary behavior is explicit. The polynomial is continuous, so EVT gives a global maximum. Since $A'=60-4x$, the only interior stationary point is $x=15$, giving $y=30$ and area $450$ square metres. The endpoint areas are zero. This comparison establishes the absolute maximum. Alternatively, $A=450-2(x-15)^2$ proves the same conclusion without differentiation.

If fencing were required on all four sides, the constraint would be $2x+2y=60$ and the maximizing rectangle would be a square of side $15$, area $225$. A correct derivative applied to the wrong constraint solves the wrong problem perfectly.

An open box

Cut squares of side $x$ from the corners of a $12$ by $12$ centimetre sheet and fold the sides. The volume is

$$V(x)=x(12-2x)^2,\qquad 0\leq x\leq6.$$

Differentiation in factored form gives

$$V'=(12-2x)(12-6x).$$

The interior stationary point is $x=2$, giving dimensions $8$ by $8$ by $2$ and volume $128$ cubic centimetres. Both endpoints give zero volume. The factor $12-2x$ also vanishes at $6$, but that is an endpoint, not an additional interior stationary box. We have checked the entire feasible interval, including the cases that are geometrically degenerate.

Minimizing on an unbounded interval

A closed cylinder has prescribed volume $V_0>0$. With radius $r>0$ and height $h$, $\pi r^2h=V_0$ and surface area is

$$S(r)=2\pi r^2+\frac{2V_0}r.$$

The derivative vanishes when $4\pi r-2V_0/r^2=0$, or $r^3=V_0/(2\pi)$. At this radius, the constraint gives $h=2r$. Because $S''=4\pi+4V_0/r^3>0$ throughout $r>0$, the derivative is increasing and the stationary point is a global minimum. Also $S\to\infty$ as $r\to0^+$ or $r\to\infty$, ruling out an escaped finite boundary optimum. EVT alone cannot be invoked on the open unbounded domain.

The model assumes the same material cost per unit area for the side and both ends, negligible thickness, and no seam allowance. Different manufacturing costs alter the objective and may alter the optimum.

Sensitivity and discrete decisions

A continuous optimum can guide a decision constrained to integer quantities, but rounding must respect the feasible set. Evaluate neighboring feasible integers, not just the nearest decimal in isolation. If an objective is unimodal, those nearby candidates often suffice; justify that structure before excluding distant alternatives.

Dimensions and parameters also matter. For fencing length $P$, the wall problem gives $x=P/4$, $y=P/2$, and maximum area $P^2/8$. Doubling available fencing quadruples the maximum area in this model. A parameterized result conveys more than one isolated numerical answer.

Exercises

1. Maximize the area of a four-sided rectangle with perimeter $80$ metres.
2. Repeat the three-sided wall problem for fencing length $P>0$.
3. For a square sheet of side $L>0$, determine the cut that maximizes the open-box volume.
4. A cylinder must have volume $16\pi$ cubic centimetres. Find the minimum-area radius and height.
5. Explain why finding $A'(x)=0$ does not by itself prove an absolute maximum.
6. Identify an assumption that would change the cylinder’s objective without changing its volume constraint.

Worked solutions

1. Write $A=x(40-x)$ on $[0,40]$. The maximum is $400$ square metres at $x=y=20$.
2. $A=x(P-2x)$; its stationary point is $P/4$, giving $y=P/2$ and area $P^2/8$. Endpoint areas vanish.
3. $V=x(L-2x)^2$ on $[0,L/2]$, with derivative $(L-2x)(L-6x)$. The maximum occurs at $x=L/6$, giving volume $2L^3/27$.
4. $r^3=16\pi/(2\pi)=8$, so $r=2$ and $h=4$ centimetres.
5. Stationary points can be minima or neither; boundaries, nondifferentiable points, and nonattainment must be checked.
6. More expensive end material, seam allowance, a missing lid, or thickness-dependent costs would change the surface-cost expression.