Why the Feasible Domain Comes First
In an optimization problem, a formula alone does not tell you which inputs are allowed. A variable might represent a length, so negative values are physically impossible. A second dimension might be determined by an equality constraint and must also be nonnegative. The objective formula can impose further restrictions: a denominator cannot be zero, and a real square root requires a nonnegative radicand.
The feasible domain is the set of values of the chosen decision variable that satisfy all the physical constraints and for which the objective formula is defined. Finding it means translating every restriction into a condition on that variable, combining the conditions, and checking whether each boundary is allowed. Do this before studying how the objective behaves.
Turn the Description into Conditions
Start with the meaning and units of the variable. If \(x\) is a length or amount that may be zero, its physical condition is \(x\geq 0\). If the model requires a positive, nonzero size, use \(x>0\) instead. These conditions are not interchangeable: the first allows a zero-size boundary case, while the second rules it out.
Next, include every other quantity that depends on \(x\). If an equality constraint gives another quantity as \(y=12-2x\), and \(y\) must be nonnegative, then \(12-2x\geq 0\), so \(x\leq 6\). A restriction on a dependent quantity can therefore set a boundary for \(x\).
Finally, check the formula itself. For real-valued expressions, an even root such as a square root requires its radicand to be at least zero. A denominator must be nonzero. If a formula includes both, both conditions apply. The formula’s requirements are additional to—not replacements for—the physical conditions.
A Reliable Procedure
Record the physical meaning of \(x\), constraints on dependent quantities, and any conditions required by the formula.
Solve inequalities carefully, keeping track of whether equality is allowed.
Keep only values satisfying all of them; this may leave more than one interval.
Substitute it into the physical conditions and the objective formula, not just into a simplified interval description.
Worked Examples
Worked Example: A Radical and a Forbidden Denominator Value
Suppose \(x\) is a length in metres constrained by \(0\leq x\leq 4\), and the objective formula is \(F(x)=\frac{\sqrt{16-x^2}}{x-1}\). Find its feasible domain.
The physical restriction gives \(0\leq x\leq 4\). The square root requires
Factoring gives \((4-x)(4+x)\geq 0\), which is equivalent to \(-4\leq x\leq 4\). Intersecting this with the physical interval still gives \(0\leq x\leq 4\). The denominator requires \(x-1\ne 0\), so \(x\ne 1\). Therefore the feasible domain is
Check the boundaries: at \(x=0\), the radical is \(\sqrt{16}=4\) and the denominator is \(-1\), so the formula is defined. At \(x=4\), the radical is \(\sqrt{0}=0\) and the denominator is \(3\), so this endpoint is also allowed. At \(x=1\), the denominator is zero, so that value is excluded. The physical and radical endpoints are included; the denominator restriction creates a gap.
As a direct check, \(F(0)=-4\) and \(F(4)=0\), both defined, whereas \(F(1)\) would require division by zero.
Worked Example: A Dependent Nonnegative Quantity
A design has two widths, in metres, related by \(y=10-2x\). Both widths must be nonnegative, and the objective expression is \(G(x)=\sqrt{x(5-x)}\). Find the feasible domain.
The first width condition is \(x\geq 0\). The second is \(y\geq 0\), so
The radical condition is \(x(5-x)\geq 0\). Its zeros are \(0\) and \(5\). For \(0\leq x\leq 5\), both factors are nonnegative, so their product is nonnegative. For \(x<0\), the first factor is negative and the second positive, making the product negative; for \(x>5\), the first is positive and the second negative. Thus the radical condition is exactly \(0\leq x\leq 5\).
Combining all conditions gives the feasible domain \([0,5]\). At \(x=0\), \(y=10\) and the radicand is zero, so the formula is defined. At \(x=5\), \(y=0\) and the radicand is also zero, so that endpoint is defined too. For a check at an interior value, \(x=2\) gives \(y=6\) and \(G(2)=\sqrt{2(3)}=\sqrt{6}\), a real number. This verifies that the derived interval is consistent with the constraints and formula.
Worked Example: Combining a Bound with a Denominator Restriction
Let \(x\) be a positive length in metres. A dependent length is \(y=\frac{12}{x}\), and the design requires \(y\leq 6\) and \(x\leq 8\). The objective expression is \(H(x)=\frac{\sqrt{x-2}}{8-x}\). Find the feasible domain.
Since \(x\) is positive, \(x>0\). The condition on \(y\) is
Because \(x>0\), multiplying both sides by \(x\) preserves the inequality: \(12\leq 6x\), so \(x\geq 2\). The independent upper bound gives \(x\leq 8\). The square root requires \(x-2\geq 0\), which is the same lower bound \(x\geq 2\). The denominator requires \(8-x\ne 0\), so \(x\ne 8\). Together these conditions give
Check the lower endpoint: at \(x=2\), \(y=12/2=6\), the radicand is zero, and the denominator is \(6\), so \(H(2)=0\) is defined. The upper boundary \(x=8\) satisfies the physical upper bound, but it makes the denominator zero and is therefore not in the feasible domain. For an interior check, \(x=4\) gives \(y=3\), a nonnegative radicand \(4-2=2\), and denominator \(8-4=4\), so \(H(4)=\sqrt{2}/4\) is defined.
Why Combining Conditions Works
Each condition describes a set of allowed values. For example, \(x\geq 0\) describes \([0,\infty)\), while \(x\ne 1\) describes \((-\infty,1)\cup(1,\infty)\). A value is feasible precisely when it belongs to every one of these sets. In set language, the feasible domain is their intersection.
This gives a short justification for the procedure: if a value satisfies all the conditions, it is physically allowed and makes the formula defined, so it belongs to the feasible domain. Conversely, any feasible value must satisfy each physical and formula condition, so it belongs to their intersection. The intersection therefore contains exactly the feasible values—neither too many nor too few.
It also explains why constraints can remove an endpoint that another condition allows. For instance, \(x\leq 8\) includes \(8\), but \(8-x\ne 0\) excludes it. The combined domain must respect both conditions.
Common Mistake: Assuming Every Physical Boundary Is Included
A frequent error is to solve the physical inequalities, write a closed interval, and stop. Consider \(0\leq x\leq 3\) with the expression \(J(x)=\frac{\sqrt{x}}{3-x}\). The physical interval includes \(x=3\), and the square root allows every \(x\) in that interval. But at \(x=3\), the denominator is zero. The correct domain is \(0\leq x<3\), not \([0,3]\).
The reverse mistake is also possible: excluding a boundary just because the objective contains a radical. A square root is defined when its radicand equals zero. In the second example, both endpoints give radicand zero, so the radical does not exclude them. Always test the equality case rather than guessing.
What the Domain Does—and Does Not—Tell You
The feasible domain identifies where an optimization problem is allowed to be evaluated. It does not say which feasible value gives the largest or smallest objective, whether an optimum exists, or whether an excluded boundary can be approached by feasible inputs. Those are separate questions. At this stage, the task is to represent the allowed inputs accurately and retain the distinction between included and excluded boundaries.
When several constraints are present, it is useful to keep a written list before combining them. In particular, do not multiply an inequality by an expression of unknown sign: the inequality direction could change. In the third example, multiplying by \(x\) was valid because the physical condition had already established \(x>0\). This sign check is part of deriving the domain correctly.
Check Your Understanding
For each question, identify the conditions and state whether the stated boundary values are feasible.
- If \(x\) is a nonnegative length and the formula contains \(\sqrt{9-x^2}\), what interval is allowed before any other restrictions are applied?
- For \(K(x)=\frac{\sqrt{x+1}}{x-2}\) with physical restriction \(0\leq x\leq 4\), which values are excluded by the formula?
- A dependent quantity is \(y=15-3x\), with \(x\geq 0\) and \(y\geq 0\). What interval follows from the physical constraints alone?
- Can a zero radicand make an objective undefined? Explain using the real square-root requirement.
- If a physical constraint allows \(x=5\), but the objective has denominator \(x-5\), is \(5\) in the feasible domain? Why?
