Questions: Optimization Problems

Question 1

A farmer wants to maximize the area of a rectangular pen using 100 meters of fencing. After differentiating and finding the critical point x = 25, what must you still do before concluding this is the maximum?

• A. Accept x = 25 as the maximum since critical points always give maxima
• B. Confirm with the second derivative test (or first derivative test) that it is a local max, and compare with endpoint values if the domain is closed
• C. Check that x = 25 is in the domain, but endpoints never need checking in optimization
• D. Differentiate the constraint equation instead of the objective function

Answer: (B)

Critical points are candidates for extrema, not guarantees. For a closed domain (e.g., 0 ≤ x ≤ 50 here), you must compare the critical point value with values at both endpoints — the absolute maximum may occur at a boundary. The second derivative test confirms whether the critical point is a local max or min.

Question 2

Once you find a critical point of your objective function and the problem asks for a maximum, you can be certain the critical point gives the maximum without further verification.

• T. True
• F. False

Answer: False

Critical points include local maxima, local minima, and saddle points. The problem asking for a maximum does not guarantee the critical point delivers one — you must verify with the first or second derivative test. On a closed domain you also need to compare with endpoint values, since the absolute maximum may occur at the boundary rather than at the critical point.

Question 3

Why must you reduce the objective function to one variable before differentiating in an optimization problem?

Model answer: Derivatives only eliminate ambiguity for single-variable functions. The constraint equation relates the variables, so substituting it eliminates all but one, producing a function whose critical points can be found by setting the derivative to zero.

Setting f'(x) = 0 is only meaningful for functions of one variable. When an optimization problem involves two quantities (say area in terms of length and width), the constraint (like a fixed perimeter) provides a relationship between them. Substituting collapses the two-variable relationship into a single-variable function. Without this step you would need multivariable calculus.