Why Optimization Needs a Complete Candidate Test
Finding a value of the variable where the derivative is zero is only one part of an optimization problem. A maximum or minimum might occur at an endpoint of the feasible domain, or at an interior point where the derivative does not exist. A complete justification must identify every relevant candidate, evaluate the objective at each one, and compare the resulting values.
The method begins before differentiation: describe precisely which values of the variable are allowed. That feasible domain determines which points can be considered and whether an endpoint is part of the problem.
Why These Points Are the Candidates
Two facts explain the candidate test. First, if a function is continuous on a closed, bounded interval, the Extreme Value Theorem guarantees that it attains an absolute maximum and an absolute minimum there. Second, an interior extremum at which the function is differentiable must have derivative zero. Thus an absolute extremum must occur at a boundary point, at an interior point where the derivative is zero, or at an interior point where the derivative does not exist.
The second statement is Fermat’s theorem. To see why it holds at a local maximum, take a sufficiently small \(h>0\). The difference quotient to the right satisfies \(\frac{f(c+h)-f(c)}{h}\leq 0\), because \(f(c+h)\leq f(c)\). For a sufficiently small \(h<0\), the same quotient is nonnegative: the numerator is nonpositive and the denominator is negative. If \(f'(c)\) exists, both one-sided limits equal that derivative. It must therefore be both nonpositive and nonnegative, so \(f'(c)=0\). At a local minimum, the inequalities reverse and the same conclusion follows.
This explains why derivative-zero points are candidates, not automatic answers. Fermat’s theorem says an interior extremum that is differentiable must have derivative zero; it does not say every point where the derivative is zero is an extremum. And the theorem does not apply at a point where the derivative does not exist, so those interior points must also be checked.
A Reliable Structure for the Justification
For a one-variable problem with feasible domain \([a,b]\), organize the work so that a reader can verify both the domain and the comparison. State why the existence theorem applies, find the interior critical points, include both boundaries, and evaluate the objective at every candidate.
Translate the constraints into an interval or another explicit set, and state which boundary points are included.
For the standard closed-interval test, establish that the objective is continuous on the closed, bounded feasible interval.
Within the feasible interval, solve \(f'(x)=0\) and check where \(f'(x)\) does not exist, while confirming that \(f\) itself is defined there.
Evaluate the objective at every included endpoint. Endpoints are candidates even if the derivative there is nonzero or undefined.
Use the objective values to identify the absolute maximum or minimum, and state the variable value where it occurs.
If the domain is not a closed, bounded interval, this procedure may need adjustment. For example, an open endpoint is not a feasible candidate, and an unbounded domain has no finite endpoint to test. In those situations the Extreme Value Theorem stated above does not guarantee an absolute best value. The candidate list must reflect the actual domain, and a separate argument may be needed to determine whether a best value exists.
Worked Example: Absolute Extrema on a Closed Interval
Worked Example: Absolute Extrema on a Closed Interval
Find the absolute maximum and minimum of \(f(x)=x^3-3x\) on \([-2,2]\), and justify the result with a complete candidate test.
1. State the domain and existence condition. The feasible domain is the closed interval \([-2,2]\). Since \(f\) is a polynomial, it is continuous there. The Extreme Value Theorem guarantees an absolute maximum and an absolute minimum.
2. Find the interior critical points. Differentiate: \(f'(x)=3x^2-3\). Set the derivative equal to zero: \(3x^2-3=0\), so \(3x^2=3\), \(x^2=1\), and \(x=-1\) or \(x=1\). The derivative exists at every real number, so there are no other interior critical points. Both \(-1\) and \(1\) lie in \((-2,2)\).
3. List and evaluate all candidates. Include the two endpoints and both interior critical points:
4. Compare and conclude. The greatest candidate value is \(2\), occurring at \(x=-1\) and \(x=2\). The least candidate value is \(-2\), occurring at \(x=-2\) and \(x=1\). Because an absolute extremum exists and every possible location for it is in the candidate list, these comparisons establish the absolute extrema on the interval.
Worked Example: Turning a Constraint into a Feasible Interval
Worked Example: Maximizing the Area of a Rectangle
A rectangle has perimeter \(20\) meters. What dimensions give the greatest area? Include degenerate endpoint cases in the candidate test.
1. Choose a variable and translate the constraint. Let \(x\) be one side length, in meters, and \(y\) the other. The perimeter constraint is \(2x+2y=20\), so \(x+y=10\) and \(y=10-x\). Nonnegative side lengths require \(x\geq 0\) and \(10-x\geq 0\). Thus the feasible interval is \(0\leq x\leq 10\). Its endpoints represent degenerate rectangles with zero area.
2. Write the objective and verify existence. The area, in square meters, is \(A(x)=xy=x(10-x)=10x-x^2\). This polynomial is continuous on the closed interval \([0,10]\), so the Extreme Value Theorem guarantees an absolute maximum and minimum.
3. Find interior critical points. Differentiate: \(A'(x)=10-2x\). Solve \(10-2x=0\), giving \(2x=10\) and \(x=5\). The derivative exists throughout the interval, so \(x=5\) is the only interior critical point.
4. Evaluate every candidate. The candidates are \(x=0,5,10\). Their areas are \(A(0)=10(0)-0^2=0\), \(A(5)=10(5)-5^2=50-25=25\), and \(A(10)=10(10)-10^2=100-100=0\). The greatest value is \(25\) square meters, at \(x=5\). The other side is \(y=10-5=5\) meters.
5. Check the result against the constraint. The proposed dimensions have perimeter \(2(5)+2(5)=20\) meters, as required, and area \(5\cdot 5=25\) square meters, matching the objective value. Therefore a \(5\)-meter by \(5\)-meter rectangle has the greatest feasible area.
Worked Example: Include an Interior Point Where the Derivative Fails
Worked Example: A Continuous Objective with a Nondifferentiable Point
Find the absolute maximum and minimum of \(g(x)=(x-1)^{2/3}\) on \([0,3]\).
1. Check the domain and existence. The function is defined and continuous for all real \(x\), so it is continuous on the closed interval \([0,3]\). The Extreme Value Theorem guarantees both absolute extrema exist.
2. Check for interior critical points. For \(x\ne 1\), the derivative is \(g'(x)=\frac{2}{3}(x-1)^{-1/3}\). This derivative is never zero where it exists. At \(x=1\), it does not exist because the denominator is zero, although \(g(1)=0\) is defined. Since \(1\) is in the interior of \([0,3]\), it must be included as a critical point.
3. Evaluate every candidate. The candidate list is \(0,1,3\). We have \(g(0)=(-1)^{2/3}=1\), \(g(1)=0^{2/3}=0\), and \(g(3)=2^{2/3}\). Since \(2^{2/3}>1>0\), the greatest value is \(2^{2/3}\) at \(x=3\), while the least value is \(0\) at \(x=1\).
4. Check the conclusions. The stated locations are feasible: \(1\) and \(3\) both belong to \([0,3]\). The objective values at those locations are \(g(1)=0\) and \(g(3)=2^{2/3}\), as reported. The comparison includes the endpoints as well as the nondifferentiable interior point, so no candidate has been omitted.
Common Mistake: Stopping at the Critical Points
A common error is to solve \(f'(x)=0\) and treat the resulting points as the complete answer. For example, consider \(f(x)=x^3\) on \([-1,2]\). Its derivative is \(f'(x)=3x^2\), so the only interior point where \(f'(x)=0\) is \(x=0\). A flawed conclusion might be, “The maximum occurs at \(x=0\) because it is the critical point.” But \(f(0)=0\), while the endpoint values are \(f(-1)=-1\) and \(f(2)=8\). The actual absolute maximum is \(8\) at \(x=2\), and the absolute minimum is \(-1\) at \(x=-1\). The critical point alone identifies neither answer.
The correction is to evaluate the objective at every candidate: \(x=-1,0,2\). The largest of \(-1,0,8\) is \(8\), and the smallest is \(-1\). This example also shows why \(f'(c)=0\) is a necessary condition for an interior extremum when the derivative exists, not a guarantee that an extremum occurs there.
What Makes the Written Justification Complete?
A well-supported conclusion connects the domain, the candidate list, and the comparison. It does not merely report a number. In a written response, make clear that all evaluated points are feasible and that the list is complete under the conditions of the candidate test.
- State the constraint and feasible set. Do not differentiate before establishing which variable values are allowed.
- Give the existence reason. For a continuous objective on a closed, bounded interval, name the Extreme Value Theorem and verify its conditions.
- Account for every type of candidate. Check interior points where the derivative is zero or undefined, and check every included endpoint.
- Compare objective values. Identify the greatest or least value, not only the input where a derivative condition holds.
- Report the requested quantity clearly. If the objective represents area, distance, or another measured quantity, state its units. If the problem requests dimensions, report them as well as the optimized value.
If the candidate list is infinite, simply evaluating a finite table is not enough; further reasoning is needed to compare values. In many AP optimization problems, however, solving the derivative equation produces finitely many interior candidates, making a direct comparison practical.
Check Your Understanding
For each question, focus on whether the domain, candidate list, and conclusion are all justified.
- What points must be included in a candidate test for a continuous objective on \([a,b]\) if it is differentiable at every interior point?
- Why does an interior point where the derivative does not exist need to be checked, provided the objective itself is defined there?
- A student finds one point where \(f'(x)=0\) on a closed interval and calls it the absolute maximum. What else must the student check?
- For \(h(x)=x^2-4x\) on \([0,5]\), identify the boundary and interior critical candidates. What values of \(h\) must be compared?
- Why does continuity on a closed, bounded interval matter before using the complete candidate test to claim that an absolute maximum and minimum exist?
