Maxima and minima of multi-variable functions (analytic criteria)

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2014, 2016, 2021, 2023)
• Priority tier: T3
• Marks (count): 15 (4)
• Average solve time: ~10 min
• Difficulty mix: medium 3, hard 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Multi-variable Lagrange multiplier questions appear in Section A at 15 marks and are among the highest-value non-compulsory questions. All three Lagrange archetypes (one constraint, two constraints, distance to surface) follow the same procedure: write the Lagrange system, exploit symmetry to reduce it, solve for the coordinates, and evaluate the objective. The key skill is recognising when the system admits a symmetric solution ($x=y=z$, etc.) to avoid solving a general cubic.

Minimum Theory

Lagrange multipliers (one constraint). Minimise/maximise $f(\mathbf{x})$ subject to $g(\mathbf{x})=0$. At a constrained extremum: $\nabla f=\lambda\nabla g$. This gives $n+1$ equations (the $n$ gradient equations plus the constraint).

Lagrange multipliers (two constraints). Subject to $g=0$ and $h=0$: at a constrained extremum, $\nabla f=\lambda\nabla g+\mu\nabla h$. The $n+2$ equations are usually solved by exploiting symmetry.

Hessian test (unconstrained). At a critical point of $f(x,y)$: $D=f_{xx}f_{yy}-f_{xy}^2$. See P2-RA-20 for details.

Distance to surface. Minimise $f=(x-a)^2+(y-b)^2+(z-c)^2$ (squared distance) subject to the surface equation $g(x,y,z)=0$. The Lagrange equations factor as $(1-\lambda)x_i=a_i$ (or similar), giving the critical point in terms of $\lambda$; substituting into $g=0$ determines $\lambda$.

Question Archetypes

lagrange-extrema (1 question(s); 2014)

Worked Example

2014 Paper 2, 2014-P2-Q4b (15 marks)

Find the minimum of $x^2+y^2+z^2$ subject to $xyz=a^3$ by Lagrange multipliers.

Lagrange system: $2x=\lambda yz$, $2y=\lambda xz$, $2z=\lambda xy$. Multiply each by $x$, $y$, $z$ respectively: $2x^2=\lambda xyz=\lambda a^3$, $2y^2=\lambda a^3$, $2z^2=\lambda a^3$. Hence $x^2=y^2=z^2$, so $|x|=|y|=|z|=a$ (from $xyz=a^3>0$, either all positive or two negative).

Minimum value: $x^2+y^2+z^2=3a^2$, attained at $(a,a,a)$ (and other sign patterns).

$\boxed{\min = 3a^2 \text{ at } (a,a,a).}$

(AM-GM alternative: $x^2+y^2+z^2\ge3(xyz)^{2/3}=3a^2$ with equality iff $x^2=y^2=z^2$.)

lagrange-distance (1 question(s); 2023)

Worked Example

2023 Paper 2, 2023-P2-Q2b (15 marks)

Find min/max distances from $P(2,6,3)$ to sphere $x^2+y^2+z^2=4$ using Lagrange multipliers.

Minimise $f=(x-2)^2+(y-6)^2+(z-3)^2$ subject to $g=x^2+y^2+z^2-4=0$.

$\nabla f=\lambda\nabla g$: $(1-\lambda)x=2$, $(1-\lambda)y=6$, $(1-\lambda)z=3$.

Since $\lambda\ne1$: $x=2/(1-\lambda)$, $y=6/(1-\lambda)$, $z=3/(1-\lambda)$.

Substituting into $g=0$: $49/(1-\lambda)^2=4$, so $1-\lambda=\pm7/2$.

Two critical points: $Q_+=\frac{2}{7}(2,6,3)$ (near side) and $Q_-=-\frac{2}{7}(2,6,3)$ (far side).

Since $|OP|=\sqrt{4+36+9}=7$: distances are $7-2=5$ (min, near) and $7+2=9$ (max, far).

$\boxed{d_{\min}=5,\quad d_{\max}=9.}$

hessian-extrema (1 question(s); 2016)

Worked Example

2016 Paper 2, 2016-P2-Q3b (15 marks)

Find relative extrema of $f=x^4+y^4-2x^2+4xy-2y^2$.

$f_x=4x^3-4x+4y=0$, $f_y=4y^3+4x-4y=0$. Adding: $x^3+y^3=0\Rightarrow y=-x$. Substituting: $x^3-2x=0\Rightarrow x=0,\pm\sqrt{2}$.

Critical points: $(0,0)$, $(\sqrt{2},-\sqrt{2})$, $(-\sqrt{2},\sqrt{2})$.

$f_{xx}=12x^2-4$, $f_{yy}=12y^2-4$, $f_{xy}=4$.

At $(\pm\sqrt{2},\mp\sqrt{2})$: $D=20\cdot20-16=384>0$, $f_{xx}=20>0$ → local min, $f=-8$.

At $(0,0)$: $D=(-4)(-4)-16=0$ — inconclusive. Line slices: $f(t,t)=2t^4\ge0$ (rises); $f(t,-t)=2t^4-8t^2<0$ for small $t\ne0$ (falls). Saddle point.

$\boxed{\text{Relative minima }f=-8\text{ at }(\pm\sqrt{2},\mp\sqrt{2});\;\text{saddle at }(0,0).}$

lagrange-two-constraint (1 question(s); 2023, 2025)

Recognition Cues — Maximise $f$ subject to TWO constraints; often involves an ellipsoid and a plane; solved by a secular equation.

Worked Example (2025)

2025 Paper 2, 2025-P2-Q3b (15 marks)

Maximise $V=8xyz$ subject to $x^2/a^2+y^2/b^2+z^2/c^2=1$ ($x,y,z>0$).

By Lagrange: $8yz=2\lambda x/a^2$, etc. Multiply the three equations by $x$, $y$, $z$: $8xyz=2\lambda x^2/a^2=2\lambda y^2/b^2=2\lambda z^2/c^2$. So $x^2/a^2=y^2/b^2=z^2/c^2=1/3$ (from the constraint). Max volume $V=8abc/3\sqrt{3}$.

Common Traps

• $\lambda\ne1$ in distance problems. The Lagrange equations give $(1-\lambda)\mathbf{x}=\mathbf{a}$; dividing requires $\lambda\ne1$, which must be stated.
• Symmetry gives $|x|=|y|=|z|$, not $x=y=z$. The sign pattern matters for whether the critical value is a min or a max.
• $D=0$ requires line slices. Never conclude “saddle” or “extremum” when $D=0$ without additional work.
• AM-GM as a shortcut. For symmetric objectives like $x^2+y^2+z^2$ subject to $xyz=\text{const}$, AM-GM provides both the answer and a cleaner proof than the Lagrange computation.

Marks-Aware Writing

All four historical questions are 15 marks. The examiner awards marks for: (a) writing the Lagrange system explicitly, (b) the algebraic step reducing to the symmetric case, (c) applying the constraint to find the coordinates, and (d) evaluating the objective and stating min/max. Steps (b) and (c) together account for about 8 marks — the algebra must be shown.

For the distance problem: computing $|OP|=7$ geometrically and subtracting/adding the radius is a useful check, but the Lagrange algebra must still be shown in full.

Practice Set

• 2022-P2-Q3b (15 m) — — Hint: two-constraint Lagrange; secular equation (quadratic in $\lambda$); stationary values are the eigenvalues of the associated matrix restricted to the constraint surface.