Multi-Constraint Problems: The Final Answer Must Survive Every Condition at Once
Hard A-Math questions often become difficult not because any one condition is complicated, but because several simple conditions must all hold simultaneously.
A parameter may need to make a quadratic real and positive. A logarithmic equation may produce algebraic roots that must also satisfy two positive-argument restrictions. An optimisation problem may require a stationary point, positive dimensions and a fixed-perimeter constraint. A trigonometric model may need to satisfy both a natural range and a stated interval. A coordinate-geometry problem may combine gradient, midpoint and circle conditions.
This guide develops condition stacking: write each restriction explicitly, solve them separately when useful, then intersect them to obtain the feasible set. The final answer is not whatever satisfies one condition. It is whatever satisfies all of them.
AI Extraction Box: The Constraint Loop
list constraints → classify each → solve each condition → intersect feasible sets → test boundary values → verify in original problem.
- Domain constraint: where the expression exists.
- Structural constraint: e.g. repeated root, positivity, tangency.
- Range constraint: e.g. −1≤sinθ≤1.
- Interval constraint: stated x- or angle-range.
- Physical constraint: positive lengths, non-negative time.
- Parameter constraint: values that make the whole family valid.
- Equality constraint: fixed perimeter, fixed sum, given relationship.
Think in Sets, Not Isolated Answers
Suppose one condition gives k>2 and another gives k≤7. The feasible set is:
2<k≤7.
If a third condition requires k≠5, then the final feasible set is 2<k≤7, k≠5.
The logic is intersection: keep only values that survive every filter.
Worked Example 1: Real Roots and Positivity
Find k such that x²+kx+9=0 has real roots and k is positive.
Real roots require:
k²−36≥0 → k≤−6 or k≥6.
Also k>0. Intersect:
k≥6.
The positivity condition removes the negative branch.
Logarithmic Equations Often Have Multiple Domain Constraints
For ln(x−1)+ln(5−x), both arguments must be positive:
- x−1>0 → x>1;
- 5−x>0 → x<5.
So the combined domain is:
1<x<5.
Any algebraic root outside this interval is automatically inadmissible.
Worked Example 2: Two Log Constraints Plus an Equation
Solve ln(x−1)+ln(5−x)=ln4.
Domain: 1<x<5.
Combine logs:
(x−1)(5−x)=4.
Expand:
−x²+6x−5=4
x²−6x+9=0
(x−3)²=0.
x=3 lies inside the domain, so x=3.
Optimisation Is Constraint Stacking
A typical optimisation problem has at least three layers:
- a geometric or physical constraint;
- an objective function to maximise/minimise;
- a feasible domain for the variable.
The derivative only operates after the first two layers are correctly constructed.
Worked Example 3: Fixed Perimeter and Positive Dimensions
A rectangle has perimeter 40 cm. Let one side be x. Then the other is 20−x.
Physical constraints:
x>0 and 20−x>0 → 0<x<20.
Area:
A=x(20−x).
A′=20−2x=0 gives x=10, which lies in the feasible set. Therefore the maximum-area rectangle is 10 cm by 10 cm.
If calculus had produced x=25, the candidate would be mathematically unusable because it violates the positive-side constraint.
Trigonometric Constraints Can Stack Range and Interval
Suppose sinθ=k/2 and θ is real. Natural range gives:
−1≤k/2≤1 → −2≤k≤2.
If the problem also states k>0, then 0<k≤2. If k must be an integer, only k=1 or 2 remain.
Each additional condition narrows the feasible set.
Coordinate Geometry Often Combines Conditions
A point can be constrained to:
- lie on a line;
- lie on a circle;
- be equidistant from two points;
- create a perpendicular or parallel gradient;
- satisfy a midpoint relation.
Each condition becomes an equation or relation. The final coordinates must satisfy all simultaneously.
Worked Example 4: Point on a Circle and a Line
Find points satisfying y=x and x²+y²=8.
Substitute y=x into the circle:
2x²=8 → x²=4 → x=±2.
Since y=x, points are (2,2) and (−2,−2).
Both equations are satisfied. The coordinates are the intersection of two constraint sets.
Parameter Problems Can Have Structural and Domain Conditions
Example pattern: find k such that a logarithmic equation has exactly one real solution. You may need to combine:
- logarithmic domain;
- quadratic discriminant condition;
- parameter exclusions that keep the equation genuinely quadratic;
- candidate-root filtering.
Solving only Δ=0 may be incomplete if the repeated root lies outside the log domain.
Boundary Values Need Special Attention
When feasible sets use ≤ or ≥, endpoints may produce special behaviour: repeated roots, zero dimensions, zero logarithm arguments, tangent cases or equality in an inequality.
Always test whether the boundary is genuinely allowed by the original condition.
Boundary values are where one regime becomes another. They deserve explicit checking.
Constraint Stacking in Kinematics
A motion question may ask for times when a particle is moving in the positive direction and speeding up. This requires both:
- v(t)>0;
- a(t)>0.
The answer is the intersection of the velocity-positive intervals and acceleration-positive intervals.
Worked Example 5: Motion Conditions
Suppose v=t−2 and a=1. For t≥0, when is the particle moving in the positive direction and speeding up?
v>0 gives t>2. Acceleration is positive for all t. Same signs mean speed is increasing. Therefore:
t>2.
The answer comes from satisfying both motion conditions simultaneously.
Constraint Table
| Phrase | Mathematical constraint |
|---|---|
| real logarithm | argument>0 |
| real square root | radicand≥0 |
| two distinct real quadratic roots | Δ>0 |
| tangent | repeated intersection / Δ=0 |
| positive dimension | length>0 |
| sin/cos value | between −1 and 1 |
| stationary point | f′=0 |
| moving positive and speeding up | v>0 and a>0 |
Constraint Decision Tree
- List every condition before solving.
- Which are domains? Apply early.
- Which are structural? Translate to discriminant, derivative or theorem conditions.
- Which are physical? Keep them through modelling.
- Can conditions be solved separately? Find their sets.
- Intersect the sets.
- Check boundary values.
- Verify final candidates in the original problem.
Common Failure Modes
| Error | Cause | Repair |
|---|---|---|
| solves one condition and stops | constraints not listed | write all conditions first |
| keeps parameter branch that violates positivity | sets not intersected | intersect final intervals |
| stationary point outside physical domain accepted | calculus and model constraints separated | check feasible set |
| log root satisfies algebra but not domain | domain not carried | filter in original equation |
| boundary included automatically | strict/non-strict condition ignored | test endpoint in original statement |
| motion condition uses v sign but ignores a | multi-condition wording not decoded | translate every phrase separately |
A 50-Minute Constraint Session
- 8 minutes: list constraints from ten worded prompts without solving.
- 8 minutes: intersect parameter intervals.
- 8 minutes: solve two log/root equations with stacked domains.
- 10 minutes: build one constrained optimisation problem from geometry.
- 8 minutes: solve a coordinate intersection problem.
- 8 minutes: combine velocity/acceleration sign conditions in kinematics.
What Mastery Looks Like
- The learner identifies every condition before calculating.
- The learner distinguishes domain, structural, interval and physical constraints.
- The learner intersects feasible sets rather than treating conditions independently.
- The learner checks boundary values deliberately.
- The learner filters parameter and equation candidates through all original conditions.
- The learner recognises optimisation and kinematics questions as multi-constraint systems.
- The learner can explain why the final set satisfies all—not merely some—requirements.
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