MTH2010 Chap.5 Maxima, Minima and Lagrange Multipliers
Maxima, Minima and Lagrange Multipliers
Optimisation in two variables is the subject of Week 5 and a frequent source of mid-semester and final examination questions. Local maxima and minima can occur only at critical points, defined in this unit as points where the gradient is zero or does not exist.
The second derivatives test classifies critical points using A = f_xx, B = f_xy and C = f_yy: if AC - B² is positive the point is a local minimum or maximum according to the sign of A, if it is negative the point is a saddle, and if it is zero the test is inconclusive. Absolute extrema are a separate question.
The extreme value theorem guarantees that a continuous function on a closed and bounded set attains both an absolute maximum and an absolute minimum, and the slides show that dropping either condition can destroy that guarantee.
The working method is a four-step routine: find interior critical points, find critical points of the function restricted to each piece of the boundary, include corner points where the boundary is not smooth, and compare all the values. Constrained problems use Lagrange multipliers.
At an extremum of f subject to g = 0, the gradients of f and g are parallel, so the system grad f = lambda grad g together with the constraint produces every candidate. With two constraints a second multiplier is added. The method finds absolute extrema directly when the constraint curve is closed and bounded; otherwise the ends of the curve must be considered as well.
What this chapter covers
- 01
Critical points: zero gradient or no gradient
- 02
Second derivatives test: the sign of AC - B² and of A
- 03
Saddle points: why they are never extrema and can be discarded
- 04
Extreme value theorem: closed, bounded and continuous
- 05
Absolute extrema routine: interior, edges, corners, comparison
- 06
Lagrange multipliers: parallel gradients plus the constraint
- 07
Two constraints: a second multiplier for each extra condition
- 08
Unbounded constraints: checking behaviour at the ends of the curve
Worked example · free
Extremes of a linear function on a circle
- 1Gradients: grad f = (1, 1) and grad g = (2x, 2y) for g = x² + y² - 2.
- 1Lagrange system: 1 = 2 lambda x and 1 = 2 lambda y, so x = y (lambda cannot be zero).
- 1Constraint: 2x² = 2, so x = 1 or x = -1, giving the points (1, 1) and (-1, -1).
- 1Compare: f(1, 1) = 2 and f(-1, -1) = -2. The circle is closed and bounded, so these are the absolute extremes.
Key terms
- Local minimum
- A point where the function value is no larger than at any nearby point of the domain.
- Second derivatives test
- A classification of critical points by the sign of f_xx f_yy - f_xy² and of f_xx.
- Extreme value theorem
- A theorem guaranteeing absolute extrema for a continuous function on a closed and bounded set.
- Constraint curve
- The set of points satisfying the condition g = 0 along which a function is optimised.
Maxima, Minima and Lagrange Multipliers FAQ
What counts as a critical point in this unit?
A point where both partial derivatives are zero, or where at least one of them fails to exist. The second case matters for functions such as cones and absolute values, whose extremes sit at sharp points.
What happens if the second derivatives test gives zero?
The test gives no information either way. Investigate the function directly near the point, for example along several lines through it or by completing the square, before deciding.
Why do corners of a region need separate attention?
The boundary is not a smooth curve at a corner, so restricting f to an edge and differentiating can miss it. Add every corner to the candidate list and evaluate f there before comparing.
Is Lagrange's method required, or can I parametrise the boundary instead?
Both can find boundary extrema, but practice paper IIa awarded zero for a boundary answer without multipliers when they were requested. Follow the method named in the question.
Does Lagrange's method work on an unbounded constraint?
It finds candidates, but absolute extrema are not guaranteed. Examine how the function behaves at the ends of the curve, as the seminar slides advise, before claiming a maximum or minimum.
Exam move
Separate the three question types and practise each as a routine. For classification, solve the gradient system by factorising, tabulate A, B, C and AC - B² at every point, and keep the B² term. For absolute extrema on a region, write the four steps as headings and fill them in, parametrising each edge with its own interval.
For constrained problems, set up the multiplier system, eliminate lambda by dividing equations after checking zero cases, and compare values at the end. The Week 5 homework includes a triangle whose maximum sits at a corner and a box inside an ellipsoid; both are good tests of whether the routine is complete.
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