= Feasible point
{title2=$x\in F$}
= Feasible solution
{synonym}
A <feasible point> satisfies every <optimization constraint> of an <optimization> problem. For the <budget constraint> example, the <feasible set> is exactly
$$
F=\{(x,10-x):0\leq x\leq10\}.
$$
It is the <line segment> from $(0,10)$ to $(10,0)$, a <compact set> and a <convex set>. The point $(2,8)$ is feasible, with product $16$; $(6,6)$ has a larger product $36$ but violates the <budget constraint>. Feasibility answers whether a decision is allowed; optimality asks whether any allowed decision is better. The <extreme value theorem> guarantees that this <continuous> <objective function> attains its <maximum> on this nonempty <compact set>.
Eliminating $y$ gives a one-variable <objective function>:
$$
q(x)=x(10-x)=25-(x-5)^2,\qquad0\leq x\leq10.
$$
By <completing the square>, $q(x)\leq25$, with equality exactly at $x=5$. Alternatively its <derivative> is $q'(x)=10-2x$, which vanishes at $5$; the endpoint values are both zero. <Completing the square> proves global optimality directly, whereas the <first-order optimality condition> only identifies a candidate. It also quantifies the cost of imbalance: moving to $(5+\delta,5-\delta)$ with $|\delta|\leq5$ loses exactly $\delta^2$ units of product. The <arithmetic-geometric mean inequality> gives another short certificate: $\sqrt{xy}\leq(x+y)/2=5$, with equality exactly when $x=y$.
For a <transportation problem> with capacity inequalities, feasibility requires nonnegative shipments, row totals no greater than factory capacities and column totals equal to shop demands. Feasibility alone does not prove minimal cost: a matching <transportation dual certificate with capacity inequalities> provides an optimality proof. The <feasible set> can be empty, and a finite <infimum> need not be attained even when the set is nonempty.
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