Maxima, Minima and Lagrange Multipliers, taught from zero

1 · Peaks and valleys have flat tops §14.7

Picture the surface z = f(x, y) as a landscape. A local maximum at (a, b) is a hilltop: f(a, b) ≥ f(x, y) for every point near (a, b). A local minimum is the bottom of a dip, with ≤ instead.

Now use Module 4's slicing picture. Stand on a smooth hilltop and slice the surface east–west through your feet. The slice is an ordinary curve, and you're at its top, so its slope is zero: f_x(a, b) = 0. Slice north–south: same argument, f_y(a, b) = 0. So:

A point inside the domain where both partials are zero, or where a partial fails to exist, is called a critical point. The "fails to exist" clause is there for sharp tips: the cone z = √(x² + y²) has its minimum at the origin, exactly where it has a point instead of a flat bottom, so its partials don't exist there.

Finding them is algebra. Take f(x, y) = x² + y² − 4x + 6y. Set f_x = 2x − 4 = 0 and f_y = 2y + 6 = 0: the only critical point is (2, −3). Is it a minimum? Complete the square in each letter: x² − 4x = (x − 2)² − 4 and y² + 6y = (y + 3)² − 9, so

Two squares can't be negative, so f ≥ −13 everywhere, with equality only at (2, −3). It's a minimum, and in fact the lowest point anywhere.

2 · A flat point that is neither: the saddle

Here is the twist that makes this module more than school calculus. Take z = x² − y². At the origin f_x = 2x = 0 and f_y = −2y = 0, so it's a critical point. But slice east–west and you get z = x², a valley with its bottom at the origin. Slice north–south and you get z = −y², a hill with its top at the origin. The origin is a low point in one direction and a high point in another: a saddle point, shaped like a horse's saddle or a mountain pass.

So a critical point is only a candidate. It may be a max, a min, or a saddle, and you need one more tool to tell which.

3 · The second-derivative test §14.7

In one variable, g″ > 0 at a critical point meant "curving up, so a minimum". In two variables there are three second partials to combine: f_xx (curving east), f_yy (curving north) and the mixed f_xy (how much the surface twists). The test packs them into one number:

Why the minus sign? If the east and north curvatures agree in sign, their product is positive, which suggests a bowl or a dome. But a big twist f_xy can bend the surface up along one diagonal and down along the other, and then it's a saddle after all. D > 0 says "the agreement beats the twist". On the saddle x² − y²: f_xx = 2, f_yy = −2, f_xy = 0, so D = −4 < 0: saddle, as the picture said.

A full example. f(x, y) = x³ + y³ − 3xy.

1. Partials: f_x = 3x² − 3y, f_y = 3y² − 3x. Setting both to zero gives y = x² and x = y².
2. Substitute the first into the second: x = x⁴, so x(x³ − 1) = 0, giving x = 0 or x = 1. (Factor; never divide by x, or you lose x = 0.) Critical points: (0, 0) and (1, 1).
3. Second partials: f_xx = 6x, f_yy = 6y, f_xy = −3.
4. At (0, 0): D = 0 · 0 − 9 = −9 < 0, a saddle. At (1, 1): D = 36 − 9 = 27 > 0 with f_xx = 6 > 0, a local minimum, value f(1, 1) = −1.

When D = 0: x⁴ + y⁴ and x⁴ − y⁴ both have D = 0 at the origin. The first is clearly a minimum (it's never negative), the second a saddle (positive along the x-axis, negative along the y-axis). Same D, different answers, which is exactly why the test must stay silent. Argue from the function itself.

4 · The exam question: absolute max and min on a region §14.7

A local max is only the top of its own neighbourhood. The exam usually asks something bigger: the highest and lowest value of f anywhere on a given region R. Two facts make that answerable.

• They exist. If f is continuous and R is closed (it contains its boundary) and bounded (it fits inside some big circle), then f actually reaches an absolute max and an absolute min on R. Every region in the past papers is of this kind.
• They can only hide in two places. Either inside R, and then the point is a critical point (section 1); or on the boundary of R, where section 1's argument doesn't apply because you can't walk in every direction.

So the whole job is to collect every point that could be the answer, then compare. That gives a routine you can run the same way every time:

Notice what is not in the routine: the second-derivative test. You never need to know whether an interior candidate is a max, a min or a saddle, because the comparison table decides everything. (A saddle simply loses the comparison.)

The routine, slowly, on a triangle

Find the absolute max and min of f(x, y) = x² + y² − 2x − 4y + 1 on the triangle T with corners (0, 0), (4, 0), (0, 4).

1. Interior. f_x = 2x − 2 = 0 and f_y = 2y − 4 = 0 give (1, 2). Inside T? Both coordinates are positive and 1 + 2 = 3 ≤ 4, so yes. Keep it: f(1, 2) = 1 + 4 − 2 − 8 + 1 = −4.
2. Bottom edge y = 0, 0 ≤ x ≤ 4: f = x² − 2x + 1, derivative 2x − 2 = 0 at x = 1. Candidate (1, 0), value 0.
3. Left edge x = 0, 0 ≤ y ≤ 4: f = y² − 4y + 1, derivative zero at y = 2. Candidate (0, 2), value 4 − 8 + 1 = −3.
4. Slanted edge x + y = 4: write y = 4 − x, 0 ≤ x ≤ 4, and substitute. (4 − x)² = 16 − 8x + x² and −4(4 − x) = −16 + 4x, so f = x² + 16 − 8x + x² − 2x − 16 + 4x + 1 = 2x² − 6x + 1. Derivative 4x − 6 = 0 at x = 3/2, so y = 5/2. Candidate (3/2, 5/2), value 2(9/4) − 9 + 1 = −7/2.
5. Corners: f(0, 0) = 1, f(4, 0) = 16 − 8 + 1 = 9, f(0, 4) = 16 − 16 + 1 = 1.
6. Compare in one table:

A sanity check you can often use. Completing the square, f = (x − 1)² + (y − 2)² − 4: that's "(distance from (1, 2))² − 4". So f is smallest at the point of T nearest to (1, 2), which is (1, 2) itself, and largest at the point farthest from it, which is the corner (4, 0) (distance √13, against √5 for the other two corners). Same answer, no calculus. The 2023 paper's function can be read the same way (see the notes).

5 · Walk the boundary yourself

Step 2 of the routine turns each boundary piece into a one-variable problem. The widget lets you watch that happen. Choose a region, then drag the slider to walk once around its boundary. The top picture shows where you are; the graph underneath is f along the walk, one piece after another. Its bumps and dips are step 2's candidates, and its breaks are the corners. "Show candidates" lays out the full comparison table. Two of the regions are straight from the past papers.

6 · Lagrange multipliers: best value along a curve §14.8

Sometimes the whole problem lives on a curve: "largest value of 3x + 4y on the circle x² + y² = 25". Or a boundary piece is awkward to parametrise. Lagrange's method handles both, using the gradient ∇f = ⟨f_x, f_y⟩ from Module 5, a vector that always points perpendicular to the level curves of f.

The idea is a picture. Walk along the constraint curve g(x, y) = c and watch which level curve of f you're standing on. Where the constraint crosses a level curve, you can keep walking onto higher levels, so you're not at the best point yet. At the best point, the constraint just touches a level curve without crossing it. Two curves that touch share a tangent line, so their perpendicular directions agree too. Both are gradients, so:

λ (lambda) is just the unknown stretch factor between the two parallel vectors. Written out, that's three equations for three unknowns: f_x = λg_x, f_y = λg_y, g = c.

Worked: f = 3x + 4y, g = x² + y² = 25.

1. ∇f = ⟨3, 4⟩, ∇g = ⟨2x, 2y⟩. Equations: 3 = 2λx, 4 = 2λy, x² + y² = 25.
2. λ can't be 0 (then 3 = 0), so it's safe to divide by it: x = 3/(2λ), y = 4/(2λ).
3. Into the circle: (9 + 16)/(4λ²) = 25, so λ² = 1/4 and λ = ±1/2. That gives (3, 4) and (−3, −4).
4. Compare: f(3, 4) = 25 is the max, f(−3, −4) = −25 is the min. As on the picture.

Two constraints (a curve cut out by two surfaces in space) use the same idea with two multipliers: ∇f = λ∇g + μ∇h, plus both constraint equations. It's in §14.8; the notes page has one short example.

Lagrange on a region's boundary. In an absolute-extrema question, a circular boundary piece can be handled either by parametrising (x = cos t, y = sin t) or by Lagrange with g = x² + y². Both give the same candidates; use whichever makes cleaner algebra, and the other as a check if you have time.

7 · You're ready: what to do next

The module in four sentences. Interior peaks and dips are critical points (∇f = 0 or undefined). D sorts critical points into max, min and saddle, and is silent when D = 0. Absolute extremes on a closed bounded region come from one routine: interior, each boundary piece, corners, one comparison table. Along a curve, the best points are where ∇f is parallel to ∇g. To turn it into marks:

• Work the two past-paper questions on the notes page: the diamond (2024) and the disc slice (2023), each with a full candidate table. Cover the solution, run the routine, then compare.
• Drill from the notes page's exercise table (§14.7–14.8, odd numbers, answers in the back).
• For the open-book exam, tab the notes page's routine card and trap list, and write the four step headings on every region question even before you start computing. The headings show the examiner your method.