MATH U101 · Multivariable Calculus · Module 6

Extreme Values, Saddle Points & Lagrange Multipliers

Thomas §14.7–14.8, lectures L12–L15. The last module in the mid-sem portions (Modules 1–6), and the one with the most mechanical, most repeatable exam question.

🎧 Narrated walkthroughs for this module

Critical points & the second-derivative test (6.7 min, T7 Q5–Q6) · Absolute max & min on a triangle (5.0 min, T7 Q4). Each one is sound-on and pauses for your next move.

Start here

The facts: "find the absolute maximum and minimum of f on this region" was a full question in both older past mid-sems: 15 marks in 2023, 12 marks in 2024 (both worked below, with candidate tables). It is solved by one four-step routine that doesn't change from region to region, so it rewards practice more than insight. The idea is 12th-class "check critical points and endpoints", with a boundary curve in place of the two endpoints; saddle points and Lagrange multipliers are new to everyone.

First time? Start with the Module 6 lesson (slow build, a boundary-walk widget on both past-paper regions). This page is the compressed map for revision, and the page to tab for the open-book exam.

1 · Critical points and local extrema §14.7

Local max / min at (a, b): f(a, b) is ≥ (or ≤) every value of f near (a, b).
First-derivative test: at an interior local extreme where the partials exist, f_x = f_y = 0 (both slices through the point are at a top or bottom).
Critical point: an interior point where f_x = f_y = 0, or where a partial doesn't exist (a sharp tip, like the cone √(x²+y²) at the origin).
Saddle point: a critical point that is neither: every small disc around it contains higher and lower values (x² − y² at the origin).
Solving f_x = f_y = 0: substitute one equation into the other, then factor. Never cancel a variable that could be zero (x = x⁴ means x(x³ − 1) = 0, two answers, not one).

2 · The second-derivative test §14.7

At a critical point with f_x = f_y = 0, compute D = f_xxf_yy − f_xy² there:

D	f_xx	Conclusion
> 0	> 0	local min
> 0	< 0	local max
< 0	any	saddle point
= 0	any	inconclusive: argue from f itself (x⁴ + y⁴ is a min, x⁴ − y⁴ a saddle, both with D = 0)

Reference result (derived in the lesson): x³ + y³ − 3xy has a saddle at (0, 0) (D = −9) and a local min −1 at (1, 1) (D = 27). Scope: the test classifies local behaviour. For an absolute-extrema question it is optional; the comparison table decides.

3 · Absolute max and min on a closed, bounded region §14.7

🎧 See the routine run first: the narrated walkthrough on Worked example 1 (about 5 min).

Why it works: a continuous f on a closed (boundary included) and bounded region R reaches an absolute max and min there, and each one is either an interior critical point or a boundary point. So collect every possible location, then compare.

The routine · write these four headings on the page

1 · Interior. Solve f_x = f_y = 0; keep a point only if it lies in R, and write one line saying why a rejected point is outside.
2 · Each boundary piece. Describe it with one variable and its range (segment: y = mx + c, x from … to …; circle: x = a + r cos t, y = b + r sin t, t from … to …). Substitute, differentiate, keep zeros strictly inside the range. (A circle can go to Lagrange instead, §4.)
3 · Corners. Every point where two pieces meet.
4 · Compare. One table: candidate, where it came from, value of f. Largest = absolute max, smallest = absolute min; state both value and location(s).

Before step 1, draw R and name its boundary pieces: most lost marks on these questions come from a misread region, not from the calculus.

Worked example 1 · 2024 mid-sem Q2(a) [12]The diamond

As set (verbatim)Find the absolute maximum and minimum values of the function z = f(x, y), where f(x, y) = 2x² + y − 3xy in the plane region R bounded by the lines |x| + |y| = 1.

|x| + |y| ≤ 1: a square standing on a corner. ▲ max, ▼ min, ● other candidates.

Region. In each quadrant the signs of x, y are fixed, so |x| + |y| = 1 is a straight line there: four edges, ±x ± y = 1, meeting at the corners (±1, 0), (0, ±1).
Interior. f_x = 4x − 3y, f_y = 1 − 3x. The second gives x = 1/3; then y = 4x/3 = 4/9. Inside, since 1/3 + 4/9 = 7/9 < 1. f(1/3, 4/9) = 2/9 + 4/9 − 4/9 = 2/9. (It's a saddle, D = 4·0 − 9 = −9, but it stays in the table.)
Edge x + y = 1, y = 1 − x, 0 ≤ x ≤ 1: f = 2x² + 1 − x − 3x + 3x² = 5x² − 4x + 1; 10x − 4 = 0 at x = 2/5: candidate (2/5, 3/5), f = 4/5 − 8/5 + 1 = 1/5.
Edge −x + y = 1, y = 1 + x, −1 ≤ x ≤ 0: f = 2x² + 1 + x − 3x − 3x² = −x² − 2x + 1; −2x − 2 = 0 at x = −1, which is an endpoint (a corner), so nothing new.
Edge −x − y = 1, y = −1 − x, −1 ≤ x ≤ 0: f = 2x² − 1 − x + 3x + 3x² = 5x² + 2x − 1; 10x + 2 = 0 at x = −1/5: candidate (−1/5, −4/5), f = 1/5 − 2/5 − 1 = −6/5.
Edge x − y = 1, y = x − 1, 0 ≤ x ≤ 1: f = 2x² + x − 1 − 3x² + 3x = −x² + 4x − 1; zero derivative at x = 2, outside [0, 1]: nothing new.
Corners: f(1, 0) = 2, f(0, 1) = 1, f(−1, 0) = 2, f(0, −1) = −1.

Candidate	From	f
(1/3, 4/9)	interior (saddle)	2/9
(2/5, 3/5)	edge x + y = 1	1/5
(−1/5, −4/5)	edge −x − y = 1	−6/5 ← min
(1, 0)	corner	2 ← max
(0, 1)	corner	1
(−1, 0)	corner	2 ← max
(0, −1)	corner	−1

Answer: absolute max 2, at both (1, 0) and (−1, 0); absolute min −6/5, at (−1/5, −4/5). Checked against a brute-force grid of the whole square. The same question with its opening moves: past-papers page.

Worked example 2 · 2023 mid-sem Q3 [15]The disc slice

As set (verbatim)Find the absolute maximum and absolute minimum values of f(x, y) = x² + y² + 2y + 2 on the region {(x, y) ∈ ℝ² : 0 ⩽ x ⩽ y and x² + (y − 1)² ⩽ 1}.

The region: disc centre (0, 1), radius 1, cut by x ≥ 0 and y ≥ x. Dashed: circles about (0, −1), the shortcut below.

Region. The disc has centre (0, 1), radius 1, and passes through (0, 0), (1, 1), (0, 2). The line y = x meets the circle where x² + (x − 1)² = 1, i.e. 2x² − 2x = 0: at (0, 0) and (1, 1). So three boundary pieces: the chord y = x from (0, 0) to (1, 1); the arc from (1, 1) up to (0, 2); the segment x = 0 from (0, 2) down to (0, 0).
Interior. f_x = 2x = 0, f_y = 2y + 2 = 0: the point (0, −1). It has y < 0, but the region needs y ≥ x ≥ 0, so it is not in R: no interior candidate.
Chord y = x, 0 ≤ x ≤ 1: f = 2x² + 2x + 2, derivative 4x + 2, zero at x = −1/2, outside the range. Only its ends count.
Segment x = 0, 0 ≤ y ≤ 2: f = y² + 2y + 2, derivative 2y + 2, zero at y = −1, outside. Only its ends count.
Arc. On the circle, x² + (y − 1)² = 1 expands to x² + y² = 2y, so there f = 2y + 2y + 2 = 4y + 2, which grows with y: no interior critical point on the arc, largest at its top (0, 2). (Parametrising x = cos t, y = 1 + sin t, 0 ≤ t ≤ π/2, gives the same: f = 6 + 4 sin t.)
Corners: f(0, 0) = 2, f(1, 1) = 1 + 1 + 2 + 2 = 6, f(0, 2) = 4 + 4 + 2 = 10.

Candidate	From	f
(0, −1)	critical point, outside R: discarded	—
(0, 0)	corner	2 ← min
(1, 1)	corner	6
(0, 2)	corner	10 ← max

Answer: absolute max 10 at (0, 2); absolute min 2 at (0, 0). (These match the handwritten answers on the paper, and a brute-force grid.)

The 30-second cross-check. Complete the square: f = x² + (y + 1)² + 1, i.e. 1 + (distance from (0, −1))². The point of R closest to (0, −1) is (0, 0), distance 1, so min 1 + 1 = 2; the farthest is (0, 2), distance 3, so max 1 + 9 = 10. The dashed circles in the figure are exactly those two distances. In an open-book exam, write the full routine for the marks and use this as your check.

4 · Lagrange multipliers §14.8

Intuition: at the best point on the constraint curve g = c, the level curve of f touches the constraint instead of crossing it, so their normals, the gradients (Module 5), are parallel. Formal: extreme values of f subject to g(x, y) = c (with ∇g ≠ 0) occur among the solutions of

∇f = λ ∇g, g = c ⟺ f_x = λg_x, f_y = λg_y, g = c

Three variables, one constraint: add f_z = λg_z. Then evaluate f at every solution and compare, exactly as in step 4 of the routine.

Solving the system without losing cases. Example: extremes of f = xy on the ellipse x² + 4y² = 8. Equations: y = 2λx, x = 8λy. Substitute the second into the first: y = 16λ²y, so y(1 − 16λ²) = 0. Case y = 0: then x = 8λ · 0 = 0, but (0, 0) isn't on the ellipse, so reject it (with that reason). Case λ = ±1/4: y = ±x/2, and the constraint gives 2x² = 8, x = ±2: the points (2, 1), (−2, −1) with f = 2 (max) and (2, −1), (−2, 1) with f = −2 (min).

Two constraints g = c₁, h = c₂: solve ∇f = λ∇g + μ∇h with both constraints. Example: extremes of f = z on the curve where the cylinder x² + y² = 2 meets the plane x + z = 1. ⟨0, 0, 1⟩ = λ⟨2x, 2y, 0⟩ + μ⟨1, 0, 1⟩ gives μ = 1, then 2λx = −1 (so λ ≠ 0) and 2λy = 0 (so y = 0). The cylinder gives x = ±√2, and the plane z = 1 − x: max 1 + √2 at (−√2, 0, 1 + √2), min 1 − √2 at (√2, 0, 1 − √2).

Worked example 3A disk: interior routine + Lagrange on the circle, checked by parametrising

Find the absolute max and min of f(x, y) = x² + 2y² − x on the closed disk x² + y² ≤ 1.

One smooth boundary piece, so no corners. ▲ the two maxima, ▼ the minimum.

Interior. f_x = 2x − 1 = 0, f_y = 4y = 0: (1/2, 0), inside (1/4 < 1). f = 1/4 − 1/2 = −1/4.
Boundary by Lagrange, g = x² + y² = 1: 2x − 1 = 2λx and 4y = 2λy. The second is 2y(2 − λ) = 0, so either y = 0 or λ = 2.
Case y = 0: the circle gives x = ±1. f(1, 0) = 0, f(−1, 0) = 1 + 1 = 2. Case λ = 2: 2x − 1 = 4x, so x = −1/2 and y² = 3/4: (−1/2, ±√3/2), f = 1/4 + 2·(3/4) + 1/2 = 9/4.
Check by parametrising x = cos t, y = sin t: f = cos²t + 2 sin²t − cos t = 2 − cos²t − cos t (using sin² = 1 − cos²). With c = cos t ∈ [−1, 1]: 2 − c² − c has derivative −2c − 1 = 0 at c = −1/2, value 9/4; ends c = 1: 0, c = −1: 2. Same candidates ✓.

Candidate	From	f
(1/2, 0)	interior	−1/4 ← min
(1, 0)	boundary, case y = 0	0
(−1, 0)	boundary, case y = 0	2
(−1/2, ±√3/2)	boundary, case λ = 2	9/4 ← max

Answer: max 9/4 at (−1/2, ±√3/2), min −1/4 at (1/2, 0).

Classic traps · the module's mark-losers

1) Forgetting the corners, the most common single loss: in both past papers the maximum sits at a corner. 2) Keeping a critical point that is outside R: 2023's (0, −1) would give a false minimum of −1. Check membership and write the reason. 3) Reporting a saddle (D < 0) as a local max or min, or using D to "decide" an absolute extreme. The table decides. 4) Dropping the range when parametrising a boundary piece: a zero of the one-variable derivative outside its interval is not a candidate (2024's x = 2, 2023's x = −1/2). 5) Dividing by a variable in a Lagrange system: split into cases instead, and reject a case only by showing it breaks the constraint. 6) No final table, or giving only the value when the question wants the point too. The table is also what earns method marks when an arithmetic slip creeps in. 7) Misreading the region: draw it, find where its pieces meet, before any calculus.

Your minimal prerequisite kit for this module

The one-variable closed-interval method: for g on [a, b], compare g at the zeros of g′ inside the interval and at a, b. Step 2 of the routine is this, once per edge.
Partial derivatives, first and second (Module 4), and the gradient ∇f = ⟨f_x, f_y⟩ (Module 5).
Solving 2×2 systems: linear ones by substitution; non-linear ones by substituting and factoring, keeping every factor's case.
Describing a boundary: a segment as y = mx + c with its x-range; a circle with centre (a, b) and radius r as (a + r cos t, b + r sin t) with its t-range; |x| + |y| = 1 as four lines.
Completing the square, for the distance cross-check (y² + 2y = (y + 1)² − 1).
Exact fractions by hand: the answers are exact (−6/5, 9/4), and the 2023 paper banned calculators.

What to practise in Thomas

Ranges are approximate for the 15th edition; check against your copy. Prefer odd numbers (answers in the back).

Skill	Where	How many
Critical points + second-derivative test	§14.7 exercises ~1–30	6–8 odd, including one with D = 0
Absolute max/min on a closed region (triangles, rectangles, disks)	§14.7 exercises ~31–38	all odd ones: this is the exam question
Past-paper regions, from a blank page	2024 Q2(a), 2023 Q3 (worked above as Examples 1–2)	2, timed at 15 minutes each
Lagrange, one constraint, two and three variables	§14.8 exercises ~1–30	6–8 odd, at least two where a case split is needed
Lagrange, two constraints	§14.8, the two-constraint exercises near the end of the section	2 odd