Variables, objectives, and feasible sets define the problem
Objective: Why is x=3 not an acceptable answer even though f′(3)=0?
Optimization compares permitted decisions. The same formula with a different feasible set is a different problem, and a technically small objective value is irrelevant if the decision violates a constraint. Start by naming the decision variable, its allowed set, the quantity being minimized or maximized, and the units of every coefficient. The chapter’s exact result is: Adding a constant to an objective changes every objective value by the same amount and therefore leaves the set of minimizers unchanged.
The precise object is: An optimization problem has a decision variable x in a feasible set C and an objective f; “minimize f(x) subject to x∈C” asks for x*∈C with f(x*)≤f(x) for every x∈C. Read “minimize” as a comparison over every feasible choice, not as an instruction to differentiate immediately. A derivative, multiplier, or algorithm is useful only after its hypotheses have been checked.
The concrete problem is: Minimize f(x)=(x−3)²+1 subject to 0≤x≤2. Reconstruct the three worked steps, verify the reported value independently, and then test this nearby failure boundary: Changing the feasible set can change the answer even when the objective formula is untouched; feasibility is not a side note.
Adding a constant to an objective changes every objective value by the same amount and therefore leaves the set of minimizers unchanged.
For any feasible x and y, f(x)≤f(y) exactly when f(x)+c≤f(y)+c.
Thus every pairwise ordering of feasible decisions is preserved.
A point minimizes f over C exactly when it minimizes f+c over C.
Minimize f(x)=(x−3)²+1 subject to 0≤x≤2.
- The unconstrained vertex is x=3, but it is not feasible because 3>2.
- On [0,2], the distance |x−3| decreases as x moves toward the right endpoint.
- Choose x*=2 and compute f(2)=1²+1=2; every smaller feasible x is farther from 3.
Result: The unique feasible minimizer is x*=2 and the minimum value is 2.