Solving Optimization Problems

3. Visual Connection

5. Concept Development

From a model to a justified solution

After the objective has been written as a one-variable function, optimization becomes an absolute-extrema problem on a feasible domain. A complete solution must find the best input, justify that it is global, recover any other requested quantities, and interpret the result in context.

Recheck the objective and domain

Before differentiating, confirm that the function represents the quantity being maximized or minimized and that its domain enforces every physical restriction. Length, time, production, and radius are usually nonnegative; a cut from a sheet must also leave positive material.

Differentiate the reduced objective

Differentiate the one-variable objective, not the original collection of unrelated formulas. Factoring the derivative often makes its zeros and sign easier to analyze. Preserve exact expressions until the final interpretation.

Generate the complete candidate set

Interior candidates include inputs where \(f'(x)=0\) and inputs where \(f'\) is undefined while \(f\) remains defined. For a closed interval, include both endpoints as candidates even though the derivative condition does not apply there.

Filter candidates through the feasible domain

An algebraic root is not automatically a physical solution. Reject values outside the domain, values that make a dimension negative or zero when a nondegenerate object is required, and extraneous roots introduced by operations such as squaring.

Closed intervals: compare candidate values

If the objective is continuous on a closed, bounded interval, the Extreme Value Theorem guarantees absolute extrema. Evaluate the objective at every feasible critical number and endpoint; the greatest output is the absolute maximum and the least is the absolute minimum.

Open or unbounded domains

There may be no endpoints to place in a candidate table. Examine limits at excluded boundaries and at infinity, then combine that end behavior with a derivative sign analysis. For example, if a cost tends to infinity at both ends of \((0,\infty)\) and decreases before one critical number and increases after it, that critical number gives the global minimum.

Choose an appropriate justification

A candidate comparison is strongest on a closed interval. The First Derivative Test proves a local maximum when the derivative changes from positive to negative and a local minimum for the reverse change. The Second Derivative Test can classify a stationary point, but a local classification alone does not prove a global result unless the domain or concavity supplies the missing argument.

Guard against extraneous roots

When an equation containing a radical is squared, record the sign condition that existed before squaring and substitute each proposed root into the unsquared equation. This prevents a root of the transformed equation from entering the final candidate set incorrectly.

Keep exact values before rounding

Use exact radicals or fractions in derivative tests and candidate comparisons. Round only after the optimum has been established, and use enough decimal places to answer the context accurately.

Recover every requested variable

The optimizing input may represent only one dimension. Substitute it into the constraint to find the remaining dimensions, demand, time, or distance. Reporting only the critical number is incomplete when the prompt asks for a design or a maximum value.

Interpret the result with units

State what is optimized, where it occurs, and its units: for example, “The maximum area is \(1250\) square meters when the sides are \(25\) meters and \(50\) meters.” Distinguish the units of the input from those of the objective.

Common errors

Frequent errors include checking only \(f'(x)=0\), ignoring endpoints, accepting an infeasible root, using the Second Derivative Test as an unsupported claim of absolute optimality, rounding too early, and giving a number without dimensions or contextual meaning.

6. Detailed Worked Example and Error Check

Example 1: Rectangle with fixed perimeter. A rectangle has perimeter \(40\) meters. With sides \(x\) and \(20-x\),

The candidates are \(0,10,20\). Their areas are \(0,100,0\), so the global maximum is \(100\) square meters. The required dimensions are \(10\) meters by \(10\) meters.

Example 2: Pen beside a river. One hundred meters of fencing forms three sides of a rectangle. If \(x\) is each side perpendicular to the river, the parallel side is \(100-2x\).

The candidates \(0,25,50\) produce areas \(0,1250,0\). Thus the maximum area is \(1250\) square meters with perpendicular sides of \(25\) meters and a parallel side of \(50\) meters.

Example 3: Open-top box. Squares of side \(x\) are cut from a \(20\)-by-\(30\) sheet.

The derivative roots are \((25\pm5\sqrt7)/3\), but only

is feasible. Both endpoints give zero volume, so this interior candidate gives the global maximum \((10000+7000\sqrt7)/27\approx1056.31\) cubic units. The box is approximately \(3.924\) high, \(12.152\) wide, and \(22.152\) long.

Example 4: Revenue. Demand is \(n(p)=500-2p\) items at price \(p\), where \(0\le p\le250\).

The candidates are \(0,125,250\). Revenue is maximized at \(p=125\), when \(250\) items are sold and the revenue is \(\$31{,}250\).

Example 5: Closest point on a parabola. Minimize the distance from \((0,3)\) to \((x,x^2)\). It is simpler to minimize the squared distance:

The candidates are \(0\) and \(\pm\sqrt{5/2}\). Since \(Q(0)=9\), \(Q(\pm\sqrt{5/2})=11/4\), and \(Q(x)\to\infty\) as \(|x|\to\infty\), both points \((\pm\sqrt{5/2},5/2)\) are globally closest. Their distance is \(\sqrt{11}/2\).

Example 6: Minimum-material open cylinder. An open cylinder has volume \(54\pi\). From \(\pi r^2h=54\pi\), \(h=54/r^2\), so

The derivative changes from negative to positive at \(r=3\sqrt[3]{2}\). Also, \(S(r)\to\infty\) as \(r\to0^+\) or \(r\to\infty\), so this is the global minimum. The constraint gives \(h=3\sqrt[3]{2}\), and the minimum area is \(27\pi\sqrt[3]{4}\).

Example 7: Minimum travel time. A traveler runs \(x\) miles at \(8\) mph and then swims to an island at \(3\) mph:

Solving \(T'(x)=0\) gives \(x=6-6/\sqrt{55}\). Squaring during the solution also requires the original sign condition \(6-x\ge0\). Comparing this feasible candidate with the endpoints gives

The traveler should run approximately \(5.191\) miles before entering the water.