Critical points — finding, classifying: Worked examples

The scenario matrix

Every critical-point problem falls into one of these cells. The last column names the example that nails it.

#	Case class	What makes it tricky	Example
A	$D > 0, f_{xx} > 0$	clean minimum	Ex 1
B	$D > 0, f_{xx} < 0$	clean maximum	Ex 2
C	$D < 0$	saddle (sign of $f_{xx}$ irrelevant)	Ex 3
D	Mixed term $f_{x y} \neq = 0$	looks like a min, is a saddle	Ex 4
E	Several critical points	classify each separately	Ex 5
F	$D = 0$ degenerate	Hessian is silent — inspect directly	Ex 6
G	A whole line/curve of critical points	infinitely many; $\nabla f$ vanishes on a set	Ex 7
H	Constrained extremum (no free critical point)	boundary / Lagrange, not the Hessian	Ex 8
I	Word problem with units	translate → optimise → check units	Ex 9

Prerequisites we lean on: Gradient and Directional Derivatives, Hessian Matrix, Quadratic Forms and Definiteness, Lagrange Multipliers, Taylor Series in Several Variables.

$f (x, y) = x^{2} + x y + y^{2}$ (cell A) Forecast first. Both variables appear squared with $+$ signs. Guess: minimum at the origin? Hold that thought — the $x y$ term could ruin it. (Why forecast? Reading the signs before computing sets a prediction to test — if $D$ later comes out negative we'll know the $x y$ term won.)

Step 1 — gradient. $f_{x} = 2 x + y, f_{y} = x + 2 y .$ Why this step? Critical points live where the ground is flat in every direction, i.e. where $\nabla f = 0$ (parent definition).

Step 2 — solve $\nabla f = 0$ . From $2 x + y = 0$ we get $y = - 2 x$ . Substitute into $x + 2 y = 0$ : $x + 2 (- 2 x) = - 3 x = 0 \Rightarrow x = 0$ , then $y = 0$ . Single critical point $(0, 0)$ . Why this step? Both partials must vanish simultaneously; a linear system solves cleanly.

Step 3 — Hessian & discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = 1$ , so $D = f_{xx} f_{y y} - f_{x y}^{2} = (2) (2) - 1^{2} = 3 > 0.$ Why this step? $D$ decides whether it's an extremum at all (parent mnemonic: "D decides, then $f_{xx}$ describes").

Step 4 — read the type. $D > 0$ and $f_{xx} = 2 > 0 \Rightarrow$ local minimum. Why this step? $D > 0$ has already told us the point is a genuine extremum (the quadratic form is definite, same sign in every direction); now $f_{xx}$ describes which kind — positive means the surface cups upward like a bowl, so it is a minimum.

Verify. The $x y$ term was not strong enough. Sanity-check by completing the square: $x^{2} + x y + y^{2} = (x + \frac{y}{2})^{2} + \frac{3}{4} y^{2} \geq 0,$ equal to $0$ only at $(0, 0)$ . Sum of squares with positive coefficients ⇒ genuine minimum. ✓

Case B — a clean maximum

Worked example Ex 2 ·

$f (x, y) = - x^{2} - y^{2} + 4 x + 6 y$ (cell B) Forecast. Negative squares pull down in every direction, the linear terms just shift the peak off the origin. Guess: a single maximum somewhere in the first quadrant. (Why forecast? Naming the expected type and rough location now means the algebra becomes a check, not a leap of faith.)

Step 1 — gradient. $f_{x} = - 2 x + 4, f_{y} = - 2 y + 6.$ Why? Flat ground condition.

Step 2 — solve. $- 2 x + 4 = 0 \Rightarrow x = 2$ ; $- 2 y + 6 = 0 \Rightarrow y = 3$ . Critical point $(2, 3)$ . Why? Each partial is linear, so one solution each.

Step 3 — Hessian. $f_{xx} = - 2, f_{y y} = - 2, f_{x y} = 0$ , so $D = (- 2) (- 2) - 0 = 4 > 0$ . Why this step? First derivatives are zero here, so they say nothing about shape; the second derivatives (the Hessian) carry the curvature, and $D$ packages them into the single number that decides extremum-vs-saddle.

Step 4 — type. $D > 0$ and $f_{xx} = - 2 < 0 \Rightarrow$ local maximum. Why this step? Once $D > 0$ we know it is an extremum; the sign of $f_{xx}$ then describes the type — negative means the surface caps over (a dome), i.e. a maximum.

Verify. Complete the square: $f = - (x - 2)^{2} - (y - 3)^{2} + 13$ . The most $f$ can be is $13$ (both squares zero), achieved at $(2, 3)$ . Peak height $f (2, 3) = 13$ . ✓

Figure — Critical points — finding, classifying

Look at the figure above — three "clean" landscapes drawn as contour maps. Left (Ex 1): the closed rings tighten onto the magenta dot, so every direction climbs away from it — a bowl / minimum. Middle (Ex 2): identical rings but colours inverted, all directions fall away from the orange dot — a dome / maximum. Right (Ex 3, built next): the contours form a cross, not closed loops — that hyperbolic pattern is the fingerprint of a saddle, where two directions disagree.

Case C — a clean saddle

Worked example Ex 3 ·

$f (x, y) = x^{2} - 4 y^{2}$ (cell C) Forecast. Up along $x$ , down along $y$ . That mismatch is the definition of a saddle — but let's let $D$ confirm it. (Why forecast? Spotting the opposite signs on the two squares before computing lets us anticipate $D < 0$ and treat the discriminant as confirmation.)

Step 1 — gradient. $f_{x} = 2 x, f_{y} = - 8 y \Rightarrow$ critical point $(0, 0)$ . Why this step? We must first locate the flat spot: solve $\nabla f = 0$ , and $2 x = 0, - 8 y = 0$ give $(0, 0)$ .

Step 2 — Hessian. $f_{xx} = 2, f_{y y} = - 8, f_{x y} = 0$ , so $D = (2) (- 8) - 0 = - 16 < 0$ . Why this step? With the gradient zero, only curvature can classify the point; the Hessian collects the second derivatives and $D = det H$ is the sign-test we read.

Step 3 — type. $D < 0 \Rightarrow$ saddle, and here $f_{xx}$ is irrelevant — the parent warns you never to consult it once $D < 0$ . Why this step? $D < 0$ means the quadratic form is indefinite — positive in some directions, negative in others — which is precisely a saddle; no further check is needed.

Verify. Walk two paths through the origin:

along $x$ (set $y = 0$ ): $f = x^{2}$ ⇒ curves up.
along $y$ (set $x = 0$ ): $f = - 4 y^{2}$ ⇒ curves down. One path up, one down ⇒ saddle. ✓ This is the geometric meaning of an indefinite quadratic form (Quadratic Forms and Definiteness).

Case D — the mixed-term disguise

Worked example Ex 4 ·

$f (x, y) = x^{2} + y^{2} - 3 x y$ (cell D — the parent's "steel-manned mistake") Forecast. Both pure second derivatives are positive ( $f_{xx} = f_{y y} = 2$ ). The naive eye screams "minimum!". Guess whether the cross-term $- 3 x y$ is strong enough to overturn that. (Why forecast? Committing to the tempting "minimum" guess first makes the coming reversal memorable — you feel the trap spring rather than reading past it.)

Step 1 — gradient. $f_{x} = 2 x - 3 y, f_{y} = 2 y - 3 x .$ Why this step? Before we can classify anything we must find where the ground is flat, i.e. solve $\nabla f = 0$ ; that starts with computing the two partials.

Step 2 — solve. $2 x - 3 y = 0$ and $2 y - 3 x = 0$ . From the first, $x = \frac{3}{2} y$ ; plug in: $2 y - \frac{9}{2} y = - \frac{5}{2} y = 0 \Rightarrow y = 0$ , so $x = 0$ . Only critical point $(0, 0)$ . Why this step? Both partials must vanish at once; solving the pair simultaneously pins down the single candidate point.

Step 3 — discriminant. $f_{xx} = 2, f_{y y} = 2, f_{x y} = - 3$ : $D = (2) (2) - (- 3)^{2} = 4 - 9 = - 5 < 0.$ Why this step matters most here? This is exactly where "both positive ⇒ min" fails. The mixed partial $f_{x y}$ measures cross-curvature; when $f_{x y}^{2} > f_{xx} f_{y y}$ it dominates and $D$ goes negative.

Step 4 — type. $D < 0 \Rightarrow$ saddle. Why this step? $D < 0$ makes the quadratic form indefinite: the completed-square coefficients have opposite signs, so $f$ rises in some directions and falls in others. That is the definition of a saddle — and crucially the sign of $f_{xx}$ plays no role here, which is exactly why the "both positive ⇒ min" reasoning collapses.

Verify. Take the diagonal direction $y = x$ : $f = x^{2} + x^{2} - 3 x^{2} = - x^{2}$ ⇒ down. Take the anti-diagonal $y = - x$ : $f = x^{2} + x^{2} + 3 x^{2} = 5 x^{2}$ ⇒ up. Down one way, up another ⇒ saddle, despite $f_{xx} > 0$ . ✓

The figure above plots $f$ along two diagonal paths through the origin, turning the 2-D surface into two familiar 1-D curves. The magenta curve is $f$ along $y = x$ : it opens downward ( $f = - x^{2}$ ). The violet curve is $f$ along $y = - x$ : it opens upward ( $f = 5 x^{2}$ ). The two curves crossing at the orange dot — one frown, one smile — is the visual proof that a point with $f_{xx} > 0$ can still be a saddle: it's the cross-term that flips one direction.

Case E — several critical points, classify each

Worked example Ex 5 ·

$f (x, y) = x^{3} + y^{3} - 3 x y$ (cell E) Forecast. A cubic in each variable coupled by $- 3 x y$ . Cubics can give more than one flat spot. Guess: two critical points, one a saddle and one an extremum. (Why forecast? Predicting how many critical points to expect guards against stopping too early — if algebra yields only one root, we know to re-check.)

Step 1 — gradient. $f_{x} = 3 x^{2} - 3 y, f_{y} = 3 y^{2} - 3 x .$ Why this step? Critical points are exactly where $\nabla f = 0$ , so the first move must be to compute the two partial derivatives — differentiating in $x$ (holding $y$ fixed) and in $y$ (holding $x$ fixed) — before we can set them to zero and hunt for flat spots.

Step 2 — solve the system. Set both to zero: $x^{2} = y, y^{2} = x .$ Substitute $y = x^{2}$ into $y^{2} = x$ : $x^{4} = x \Rightarrow x (x^{3} - 1) = 0 \Rightarrow x = 0$ or $x = 1$ .

$x = 0 \Rightarrow y = 0$ : point $(0, 0)$ .
$x = 1 \Rightarrow y = 1$ : point $(1, 1)$ . Why substitute? It collapses two equations into one polynomial, exposing all roots.

Step 3 — Hessian (depends on the point). $f_{xx} = 6 x, f_{y y} = 6 y, f_{x y} = - 3$ , so $D = (6 x) (6 y) - (- 3)^{2} = 36 x y - 9.$ Why this step? Because $f_{xx} = 6 x$ and $f_{y y} = 6 y$ change with position, the discriminant is now a formula in $(x, y)$ rather than a fixed number — we compute it symbolically so we can plug in each critical point separately.

Step 4 — classify each cell separately.

At $(0, 0)$ : $D = 36 (0) - 9 = - 9 < 0 \Rightarrow$ saddle.
At $(1, 1)$ : $D = 36 (1) - 9 = 27 > 0$ , and $f_{xx} = 6 (1) = 6 > 0 \Rightarrow$ local minimum. Why one by one? $D$ is now a function of position; the same formula gives different verdicts at different points.

Verify. Heights: $f (0, 0) = 0$ and $f (1, 1) = 1 + 1 - 3 = - 1$ . The minimum $(1, 1)$ sits below the saddle — consistent (a valley bottom lower than a nearby pass). ✓

Case F — the $D = 0$ degenerate trap

Worked example Ex 6 ·

$f (x, y) = x^{2} y^{2}$ (cell F) Forecast. The product of two squares is $\geq 0$ . Guess: minimum everywhere it's zero? But watch $D$ collapse. (Why forecast? Anticipating that $f \geq 0$ primes us to trust direct inspection when the Hessian test — as we suspect — turns out to say nothing.)

Step 1 — gradient. $f_{x} = 2 x y^{2}, f_{y} = 2 x^{2} y .$ Why? Product rule in each variable separately.

Step 2 — solve. $2 x y^{2} = 0$ and $2 x^{2} y = 0$ . Both hold whenever $x = 0$ or $y = 0$ . So both coordinate axes are critical! Focus on the origin $(0, 0)$ (the interesting isolated-looking one; axes are Case G territory). Why this step? We are hunting every point where $\nabla f = 0$ ; factoring each partial shows the solution set is not a lone point but two whole lines, which is itself the warning that this problem is degenerate.

Step 3 — Hessian at origin. $f_{xx} = 2 y^{2}, f_{y y} = 2 x^{2}, f_{x y} = 4 x y$ . At $(0, 0)$ all three are $0$ : $D = (0) (0) - 0^{2} = 0.$ Why the test dies here. The quadratic part of the Taylor expansion is entirely zero — the second-order term carries no information. We must look at higher-order terms (Taylor Series in Several Variables).

Step 4 — inspect directly. $f = x^{2} y^{2} = (x y)^{2} \geq 0$ , and $f = 0$ on both axes. So the origin is a (non-strict) minimum — but a flat, degenerate one, not the tidy bowl of Case A. Why this step? With $D = 0$ the Hessian is silent, so we fall back on the definition of a minimum — is $f$ never smaller nearby? Since $(x y)^{2}$ can never dip below $0$ , the origin genuinely is a minimum.

Verify. Along $y = x$ : $f = x^{4} \geq 0$ ⇒ up. Along either axis: $f \equiv 0$ ⇒ flat. Never negative ⇒ minimum, confirmed by inspection, not by $D$ . ✓

Case G — a whole line of critical points

Worked example Ex 7 ·

$f (x, y) = (x - y)^{2}$ (cell G) Forecast. This is a "trough" that runs along a line. Guess: not one critical point but a whole set of them. (Why forecast? Predicting a set rather than a point stops us from wrongly declaring a single isolated minimum when the algebra hands back a whole line.)

Step 1 — gradient. $f_{x} = 2 (x - y), f_{y} = - 2 (x - y) .$ Why this step? Critical points are where $\nabla f = 0$ ; we differentiate first so we can see which points make both partials vanish — and here the chain rule pulls out the common factor $(x - y)$ .

Step 2 — solve. Both vanish whenever $x - y = 0$ , i.e. along the entire line $y = x$ . Infinitely many critical points. Why a line? The function only "sees" the combination $x - y$ ; move along $y = x$ and nothing changes.

Step 3 — Hessian. $f_{xx} = 2, f_{y y} = 2, f_{x y} = - 2$ , so $D = (2) (2) - (- 2)^{2} = 4 - 4 = 0$ everywhere. Why $D = 0$ again? A degenerate (rank-deficient) Hessian is exactly what a flat trough looks like — one eigenvalue of $H$ is zero (Hessian Matrix).

Step 4 — inspect. $f = (x - y)^{2} \geq 0$ , zero on the line $y = x$ . Perpendicular to the line the ground curves up; along the line it's flat. Each point of $y = x$ is a (non-strict) minimum. Why this step? Since $D = 0$ leaves the Hessian test mute, we return to the definition: $(x - y)^{2}$ is never negative and equals $0$ on the line, so every point of that line is a minimum.

Verify. Take the point $(1, 1)$ (on the line): $f = 0$ . Step off perpendicularly to $(1, 0)$ : $f = (1 - 0)^{2} = 1 > 0$ . Step along the line to $(2, 2)$ : $f = 0$ , unchanged. A valley floor. ✓

The figure above is the contour map of the trough. The magenta diagonal is the line $y = x$ where every point is critical — notice the contour colour is constant along it, showing the ground is dead flat in that direction. The violet arrow points across the line, where the contours bunch together and $f$ climbs as $(x - y)^{2}$ . A trough with a flat floor and rising walls: that is what an infinite line of minima looks like.

Case H — a constrained extremum (no free critical point)

Worked example Ex 8 · Maximise

$f (x, y) = x y$ on the circle $x^{2} + y^{2} = 8$ (cell H) Forecast. Freely, $f = x y$ has only a saddle at the origin (check: $f_{x} = y, f_{y} = x \Rightarrow (0, 0)$ ; $D = (0) (0) - 1^{2} = - 1 < 0$ ). So any max must live on the constraint curve. Guess the maximum sits where $∣ x ∣ = ∣ y ∣$ . (Why forecast? Realising in advance that the free critical point is a saddle tells us the answer must lie on the boundary — that steers us to Lagrange instead of the Hessian.)

Step 1 — set up Lagrange. With constraint $g = x^{2} + y^{2} - 8 = 0$ , the method of Lagrange Multipliers demands $\nabla f = λ \nabla g$ : $y = λ (2 x), x = λ (2 y) .$ Why Lagrange, not the Hessian? The Hessian test finds interior flat spots; here the extremum is forced onto a boundary curve, so the free $\nabla f = 0$ machinery does not apply.

Step 2 — solve. Multiply the two equations: $x y = 4 λ^{2} x y$ . If $x y \neq = 0$ then $4 λ^{2} = 1 \Rightarrow λ = \pm \frac{1}{2}$ . Take $λ = \frac{1}{2}$ : then $y = x$ ; feed into the constraint $x^{2} + x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$ . Points $(2, 2)$ and $(- 2, - 2)$ . Why this step? The Lagrange condition plus the constraint is a system of three equations in $x, y, λ$ ; multiplying to eliminate $λ$ first, then using the constraint to fix the size, is the cleanest route to all candidate points.

Step 3 — evaluate & compare. $f (2, 2) = 4$ , $f (- 2, - 2) = 4$ . (The $λ = - \frac{1}{2}$ branch gives $y = - x$ , points $(2, - 2), (- 2, 2)$ with $f = - 4$ , the minima.) Why compare? On a closed curve the extreme values are found by evaluating $f$ at all candidate points.

Step 4 — conclusion. Maximum value $4$ , at $(2, 2)$ and $(- 2, - 2)$ ; minimum $- 4$ at the other pair. Why this step? Optimisation asks for the actual extreme value and where it occurs; having listed all candidates, we simply pick the largest ( $4$ ) and smallest ( $- 4$ ) function values to state the answer.

Verify. Parametrise the circle: $x = 8 cos θ, y = 8 sin θ$ , so $f = 8 cos θ sin θ = 4 sin (2 θ)$ , whose max is $4$ . ✓ Matches.

Case I — a word problem with units

Worked example Ex 9 · Cheapest open-top box

(cell I) A rectangular open-top box must hold $V = 32 m^{3}$ . Material costs are uniform. Minimise the surface area $S$ (base + four sides).

Forecast. Guess the box is not a cube (no top!) — it should be wider than it is tall. (Why forecast? A physical prediction — squat, not cubic — gives us a sanity target: if the algebra returns a tall thin box we'll know a sign or constraint went wrong.)

Step 1 — model with two free variables. Let base be $x \times y$ (metres) and height $h$ . Volume fixes $h = \frac{32}{x y}$ . Open-top area: $S (x, y) = x y + 2 x h + 2 y h = x y + \frac{64}{y} + \frac{64}{x} [m^{2}] .$ Why substitute $h$ ? It reduces three variables to two, turning a constrained problem into a free critical-point problem in $(x, y)$ .

Step 2 — gradient. $S_{x} = y - \frac{64}{x ^{2}}, S_{y} = x - \frac{64}{y ^{2}} .$ Why this step? The cheapest box is a minimum of $S$ , and minima sit at critical points; so we compute $\nabla S$ in order to solve $\nabla S = 0$ .

Step 3 — solve. Set both to zero: $y = \frac{64}{x ^{2}}$ and $x = \frac{64}{y ^{2}}$ . By symmetry try $x = y$ : then $x = \frac{64}{x ^{2}} \Rightarrow x^{3} = 64 \Rightarrow x = 4$ . So $x = y = 4$ m, and $h = \frac{32}{16} = 2$ m. Why this step, and why the symmetry guess? Both partials must vanish together; the two equations are symmetric under swapping $x \leftrightarrow y$ , so the extremum is expected on the symmetric line $x = y$ , which collapses the system to a single solvable cubic.

Step 4 — confirm it's a minimum (Hessian). $S_{xx} = \frac{128}{x ^{3}}, S_{y y} = \frac{128}{y ^{3}}, S_{x y} = 1$ . At $(4, 4)$ : $S_{xx} = S_{y y} = \frac{128}{64} = 2$ , so $D = (2) (2) - 1^{2} = 3 > 0, S_{xx} = 2 > 0 \Rightarrow minimum. ✓$ Why check? A word problem needs the right kind of critical point — a genuine minimum, not a saddle.

Verify (value & units). $S (4, 4) = 16 + \frac{64}{4} + \frac{64}{4} = 16 + 16 + 16 = 48 m^{2}$ . Volume check: $4 \cdot 4 \cdot 2 = 32 m^{3}$ ✓. Units: area in $m^{2}$ ✓. The base $4 \times 4$ is wider than the height $2$ — matching the forecast (no top ⇒ short and squat). ✓

Active Recall

Recall In Ex 4, both

$f_{xx}$ and $f_{y y}$ were $+ 2$ . Why a saddle? Because $f_{x y} = - 3$ makes $D = 4 - 9 = - 5 < 0$ . The mixed term's cross-curvature dominated; $D < 0$ overrides the sign of $f_{xx}$ .

Recall When does the second derivative test go silent, and what do you do?

When $D = 0$ (Ex 6, Ex 7) — the quadratic Taylor term vanishes or is rank-deficient. Inspect the function directly or use higher-order terms.

Recall Why did Ex 8 need Lagrange multipliers instead of the Hessian?

The extremum was forced onto a constraint curve; the free $\nabla f = 0$ condition finds only interior flat points, and here the only interior critical point was a saddle.

Recall In the box problem (Ex 9), why is the optimal box not a cube?

With no top, only the base and four sides cost material, so it pays to spread out the base ( $4 \times 4$ ) and keep height low ( $2$ ), rather than equalise all edges.

Connections

Critical points — finding, classifying — the parent topic these examples exhaust.
Hessian Matrix — degenerate ( $D = 0$ ) cases are rank-deficient Hessians (Ex 6, 7).
Quadratic Forms and Definiteness — min/max/saddle ↔ definite/indefinite forms (Ex 3, 4).
Lagrange Multipliers — constrained extrema (Ex 8).
Taylor Series in Several Variables — why $D = 0$ forces higher-order inspection.
Gradient and Directional Derivatives — $\nabla f = 0$ starts every example.

Worked examples — Critical points — finding, classifying

The scenario matrix

Case A — a clean minimum

Case B — a clean maximum

Case C — a clean saddle

Case D — the mixed-term disguise

Case E — several critical points, classify each

Case F — the $D = 0$ degenerate trap

Case G — a whole line of critical points

Case H — a constrained extremum (no free critical point)

Case I — a word problem with units

Active Recall

Connections