Why Reduce a Model to One Variable?
An optimization model may begin with several quantities, even when a constraint ties them together. For instance, the length and width of a rectangle are both variables, but a fixed perimeter relates them. The objective, such as area, can then be written using just one of those variables.
This reduction matters because calculus methods for optimizing a function of one variable can be applied only after the relationship among the variables has been handled. The reduction itself is an algebraic modeling step: it must preserve the constraint and any restrictions on the variables.
Here the goal is to build and check single-variable models, not to find their maximum or minimum. A reduced expression is useful only if it represents the same feasible choices as the original constrained model.
From a Constraint to a Reduced Objective
Suppose the objective is \(Q=f(x,y)\), and the constraint is \(g(x,y)=0\). If the constraint can be solved for \(y\) as \(y=\phi(x)\), substitute that expression into the objective:
The new function \(F\) is the objective written in terms of \(x\) alone. But the substitution has a validity condition: \(x\) must be one of the values for which the original constraint and all variable restrictions are satisfied. For example, an expression involving \(1/x\) requires \(x\ne 0\), and a square root used for a length must represent a nonnegative length.
A constraint may also be solved in the other direction, as \(x=\psi(y)\). That gives another single-variable expression, \(f(\psi(y),y)\). The two expressions may look different while representing the same feasible pairs.
Why the Substitution Preserves the Model
The essential requirement is that the chosen expression for the dependent variable describes exactly the allowed pairs on the part of the constraint being modeled. If each permitted \(x\) corresponds to the pair \((x,\phi(x))\), then evaluating \(F(x)\) gives the original objective at that pair.
If \((x,y)\) satisfies the constraint and \(y=\phi(x)\), then its objective value is \(f(x,y)=f\bigl(x,\phi(x)\bigr)=F(x)\).
For an allowed input \(x\), set \(y=\phi(x)\). The expression must satisfy the original constraint and the model's restrictions.
Every represented input produces an allowed original pair, and every pair represented by the chosen branch produces the same objective value in the reduced model.
This reasoning also explains a limitation. If solving the constraint gives several possible values of \(y\) for one \(x\), a single formula may describe only one branch. To represent all allowed pairs, use separate formulas for the branches or choose a different parameter.
Worked Example: Area from a Fixed Perimeter
Worked Example: Area from a Fixed Perimeter
A rectangle has perimeter \(40\) units. Let its side lengths be \(x\) and \(y\), in units. Write its area using \(x\) alone, then using \(y\) alone, and check that the two reductions agree for a specific rectangle.
The objective is area, \(A=xy\). The perimeter constraint is
To eliminate \(y\), divide the constraint by \(2\):
Subtract \(x\) from both sides to obtain \(y=20-x\). Substitution into \(A=xy\) gives
Alternatively, subtract \(y\) from \(x+y=20\) to get \(x=20-y\). Then
These formulas use different independent variables. To check them, choose \(x=6\) units. The constraint gives \(y=20-6=14\) units. The perimeter check is \(2(6)+2(14)=12+28=40\) units. The original area is \(6\cdot14=84\) square units. The first reduced formula gives \(A(6)=6(20-6)=6\cdot14=84\) square units. The second gives \(A(14)=(20-14)14=6\cdot14=84\) square units.
Both reductions describe the same rectangle and the same area. Since the side lengths are positive, the substitutions must also represent positive side lengths; they are not instructions to treat arbitrary real values as rectangle dimensions.
Compare the Substitutions Before Calculating
Choosing which variable to eliminate is often a matter of convenience. The choice can make an expression shorter, avoid a complicated formula, or make a model's restrictions easier to state. Either way, check the algebra against the original constraint rather than judging a reduction only by how familiar it looks.
A useful check has three parts: substitute the dependent-variable expression back into the constraint, substitute it into the objective, and verify that the resulting variables satisfy any stated conditions. If dividing by a variable or taking a square root was involved, include the corresponding restriction in that check.
Worked Example: Perimeter with a Fixed Product
Worked Example: Perimeter with a Fixed Product
A rectangle has area \(36\) square units. Let its positive side lengths be \(x\) and \(y\), in units. Express the perimeter in terms of \(x\), then in terms of \(y\).
The constraint is \(xy=36\), and the perimeter objective is \(P=2x+2y\). To eliminate \(y\), divide both sides of \(xy=36\) by \(x\):
This division requires \(x\ne0\). Because \(x\) is a positive side length, the model has \(x>0\). Substitute into the perimeter:
For the other substitution, divide \(xy=36\) by \(y\). The conditions \(y\ne0\) and \(y>0\) hold because \(y\) is a positive side length. Thus \(x=36/y\), and
Check both forms using \(x=4\) units. Then \(y=36/4=9\) units, and the area check is \(4\cdot9=36\) square units. The original perimeter is \(2(4)+2(9)=8+18=26\) units. The first reduced formula gives \(P(4)=2(4)+72/4=8+18=26\) units. For the second formula, use \(y=9\): \(P(9)=72/9+2(9)=8+18=26\) units.
The two expressions differ in appearance, but they agree when \(y=36/x\). The restriction \(x>0\), or \(y>0\) in the alternative form, is part of the model—not an optional detail. In particular, the rational expression is not defined at zero.
Branch Choice Matters
Some constraints do not give just one value of a dependent variable for each independent variable. For example, solving \(x^2+y^2=25\) for \(y\) requires a choice between a positive and negative square root. If \(x\) and \(y\) represent positive lengths, the positive root is the relevant one. If the variables are signed coordinates, both branches may be allowed.
Worked Example: A Square-Root Constraint
Worked Example: A Square-Root Constraint
Suppose \(x\) and \(y\) are positive coordinates satisfying \(x^2+y^2=25\), and the objective is \(Q=x+2y\). Reduce the objective to one variable in two ways and check the results at \(x=3,\ y=4\).
First solve for \(y\). Subtract \(x^2\) from the constraint to get \(y^2=25-x^2\). Taking square roots yields two algebraic possibilities, \(y=\pm\sqrt{25-x^2}\). Since \(y\) is positive, choose the positive branch:
For this expression to be real, the radicand must satisfy \(25-x^2\ge0\); the positive-coordinate condition further requires the chosen value of \(y\) to be positive. These conditions come from the square root and the stated model.
Instead solve for \(x\). The equation \(x^2=25-y^2\) gives \(x=\pm\sqrt{25-y^2}\); because \(x\) is positive, use the positive branch. The other reduction is
Now check \(x=3,\ y=4\). The constraint is satisfied because \(3^2+4^2=9+16=25\). The original objective is \(Q=3+2(4)=3+8=11\). In the first reduction, \(Q(3)=3+2\sqrt{25-3^2}=3+2\sqrt{25-9}=3+2\sqrt{16}=3+2(4)=11\). In the second, \(Q(4)=\sqrt{25-4^2}+2(4)=\sqrt{25-16}+8=\sqrt{9}+8=3+8=11\).
The checks confirm that both formulas reproduce the original objective at the chosen point. If the positivity conditions were removed, retaining only the equation would allow negative square-root branches too. Using only the positive branch would then omit some solutions.
A Common Mistake: Treating a Dependent Variable as Constant
Consider the constraint \(x+y=10\) and objective \(R=x^2+y^2\). A common incorrect reduction substitutes \(y=10\), producing \(R=x^2+100\). This does not follow from the constraint: it sets \(y\) to a constant without accounting for the value of \(x\).
The correct algebra is \(y=10-x\), so
For a check, take \(x=3\). The constraint then gives \(y=10-3=7\), and \(3+7=10\). The original objective is \(3^2+7^2=9+49=58\). The correct reduced formula gives \(R(3)=2(3^2)-20(3)+100=18-60+100=58\). The incorrect formula gives \(3^2+100=109\), which does not match the original objective.
Why Careful Reduction Matters
After reduction, a constrained objective becomes a one-variable function that can be studied with calculus. But calculus applied to the wrong expression, the wrong square-root branch, or an expression that ignores a restriction answers a different question from the one the model asks.
Before using a reduced objective, confirm that it came from the stated constraint, that it uses the intended branch where needed, and that each operation used in solving the constraint was valid. Keep units as well: if \(x\) and \(y\) are lengths, then a perimeter has units of length and an area has units of square length.
Different choices of independent variable can be equally correct. Prefer the form that is easiest to interpret and work with, but preserve the original relationship. The key test is whether the reduced expression assigns the original objective value to every pair represented by the model.
Check Your Understanding
For each question, identify the substitution and check the conditions needed for it to represent the original model.
- Given \(x+y=18\) and objective \(A=xy\), write \(A\) as a function of \(x\) and as a function of \(y\).
- If \(xy=20\) and \(y\) is eliminated, what condition is required before dividing by \(x\)?
- For \(x^2+y^2=16\), what additional information determines whether \(y=\sqrt{16-x^2}\), \(y=-\sqrt{16-x^2}\), or both branches are appropriate?
- Explain why substituting a dependent-variable formula into the objective preserves its value for represented pairs.
- For the constraint \(x+y=8\) and objective \(x^2+y^2\), write the reduced objective using \(x\), then verify it at \(x=3\).
