ECON1202 Chap.12 Multivariable Optimisation
Multivariable Optimisation
Unconstrained optimisation begins with stationary points where both first partial derivatives are zero. These equations locate candidates but do not classify them. The course second-derivative test uses D1=fxx and D2=fxx fyy minus fxy squared. Positive D2 with negative D1 gives a local maximum; positive D2 with positive D1 gives a local minimum; negative D2 gives a saddle; zero makes the test inconclusive.
Derivatives must be evaluated at each candidate when curvature varies across the domain. Equality constraints change the candidate equations. The Lagrangian combines the objective with a multiplier times the constraint gap. Differentiation with respect to the multiplier reproduces feasibility, while the x and y conditions express a gradient relationship between objective and constraint.
Candidates must be checked in the original constraint and compared with any boundary or corner points required by the domain. Local and global claims remain distinct. The bordered-Hessian sign rule should be used with the exact arrangement printed on the course formula sheet because reordering can change determinant signs.
What this chapter covers
- 01
First-order conditions
- 02
Stationary points
- 03
Second partial derivatives
- 04
Hessian determinant test
- 05
Local maxima and minima
- 06
Saddle points
- 07
Lagrange first-order conditions
- 08
Feasibility and boundary checks
Worked example · free
Classify an unconstrained quadratic objective
- +1Set fx=-2(x-3)=0 and fy=-6(y+1)=0 to obtain (3,-1).
- +1Calculate fxx=-2, fyy=-6 and fxy=0.
- +1D2=(-2)(-6)-0^2=12>0 and D1=-2<0, so the point is a local maximum.
- +1Evaluate f(3,-1)=12; the negative sum-of-squares form independently confirms the maximum.
Key terms
- Stationary point
- A point where all first partial derivatives of an unconstrained objective are zero.
- Local maximum
- A feasible point whose objective value is at least as large as values at nearby feasible points.
- Saddle point
- A stationary point with rising directions and falling directions rather than a local extremum.
- Hessian determinant
- The curvature combination fxx fyy minus fxy squared used in the two-variable classification test.
- Lagrange multiplier
- The auxiliary variable used to join an equality constraint to an objective's first-order conditions.
- Feasible set
- The collection of variable choices that satisfy all stated constraints.
Multivariable Optimisation FAQ
Why are first-order conditions not enough?
They identify stationary candidates but do not distinguish maxima, minima or saddles. Use the second-derivative test, and state inconclusive when its determinant is zero instead of forcing a classification.
How is the two-variable second-derivative test applied?
At each stationary point calculate D1=fxx and D2=fxx fyy-fxy^2. Positive D2 uses the sign of D1 to separate minimum from maximum, negative D2 gives a saddle, and zero is inconclusive.
What does the Lagrange multiplier method add?
It keeps the equality constraint inside the first-order system. The multiplier derivative restores the constraint, while the other equations express the objective gradient as proportional to the constraint gradient at a candidate.
How do I distinguish local from global conclusions?
Derivative tests normally describe nearby behaviour. A global claim may require comparing every feasible candidate, including endpoints or corners, or using additional structure such as concavity. State only the scope supported by the work.
What evidence supports a complete multivariable optimum claim?
Start by stating the objective, variables, domain and any constraint. For an unconstrained interior problem, solve both first partial derivatives simultaneously and list every stationary point.
Compute fxx, fyy and fxy, evaluate them at each candidate, and form D2=fxx fyy-(fxy)^2. A positive determinant needs the sign of fxx to distinguish a local minimum from a local maximum; a negative determinant indicates a saddle, while zero is inconclusive and requires another argument.
For an equality constraint, build the Lagrangian, differentiate with respect to both choice variables and the multiplier, and solve the three equations together. Recheck the original constraint and compare the objective values of all feasible candidates. Consider relevant boundaries or domain restrictions separately. Keep feasibility, stationarity, local classification and global selection as distinct claims.
The final wording should not call a candidate globally optimal unless the comparison or additional shape argument actually establishes that scope. If direct substitution can eliminate one variable without losing feasible cases, use the resulting one-variable problem as an independent check. Report the objective value at every feasible candidate so the ranking is visible rather than asserted.
Exam move
Solve both first-order equations together and list every stationary point before calculating curvature. Write D1 and D2 symbolically, then substitute each candidate so location-dependent signs are not overlooked. Pair a maximum, minimum, saddle and inconclusive example to rehearse all branches of the test.
For constraints, write the Lagrangian in the same sign convention as the course formula sheet and derive all three first-order conditions, including feasibility. Substitute each candidate into the original constraint before evaluating the objective. Add endpoint or corner comparisons whenever the feasible domain requires them. Use completed-square or contour reasoning as an independent check on simple objectives.
End each solution with the variable values, objective value, feasibility statement and local or global scope so the optimisation claim is not broader than the evidence.
Working through Multivariable Optimisation in ECON1202? Sia is AskSia’s AI Economics tutor — ask any ECON1202 Multivariable Optimisation question and get a clear, step-by-step explanation grounded in how ECON1202 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
