Lagrange’s method of multipliers (constrained extrema)

At a Glance

• Frequency: 8 sub-parts across 8 of 13 years (2013, 2014, 2015, 2016, 2019, 2020, 2022, 2025)
• Priority tier: T1
• Marks (count): 13 (1), 15 (2), 20 (5)
• Average solve time: ~12 min
• Difficulty mix: hard 3, easy 3, medium 2
• Section: A | Dominant type: computation

Why This Chapter Matters

Lagrange multipliers appear in 8 of the last 13 years, predominantly as 20-mark Section A questions — each appearance can swing an attempt by two full parts. The method is universal: one procedure handles distances, volumes, applied word problems, and the abstract maximum-on-an-ellipsoid problems that UPSC particularly favours. Every question reduces to the same four-step skeleton; the only variation is recognising which type you are looking at and how many constraints are present.

Minimum Theory

The method. To find the extreme values of $f(x,y,z)$ subject to a single constraint $g(x,y,z)=0$, form the Lagrangian $L=f-\lambda g$ and set all partial derivatives to zero: $\frac{\partial f}{\partial x}=\lambda\frac{\partial g}{\partial x},\qquad \frac{\partial f}{\partial y}=\lambda\frac{\partial g}{\partial y},\qquad \frac{\partial f}{\partial z}=\lambda\frac{\partial g}{\partial z},\qquad g(x,y,z)=0.$ The geometric content: at an extremum the level surface of $f$ is tangent to the constraint surface, so their normals are parallel — $\nabla f=\lambda\nabla g$.

Two constraints. When there are two constraints $g=0$ and $h=0$, introduce two multipliers: $\nabla f=\lambda\nabla g+\mu\nabla h,\qquad g=0,\qquad h=0.$ This gives five equations for the five unknowns $(x,y,z,\lambda,\mu)$. For the specific problem of extremising $r^2=x^2+y^2+z^2$ on the intersection of a quadric and a plane, the trick is to multiply each gradient equation by its variable and sum — this identifies $\lambda=r^2$ (see quadratic-form-extrema below).

Nature of critical points. Lagrange multipliers find stationary points; max or min must be confirmed separately. On a compact constraint (ellipse, sphere) the largest value is the maximum and the smallest the minimum. On an unbounded constraint (a plane), a single stationary point of $x^2+y^2+z^2$ is always a minimum since the function grows without bound.

Question Archetypes

Four patterns cover all Lagrange multiplier questions in this atom.

constrained-distance (2 question(s); 2013, 2020)

Recognition Cues

• “Find the shortest distance between $\ldots$ and $\ldots$ by Lagrange multipliers.”
• “Find the extreme value of $u=x^2+y^2+z^2$ subject to $ax+by+cz=d$.”
• The objective is a squared distance or squared linear functional; the constraint is a curve, plane, or surface.

Solution Template

1. Write the squared distance/objective $f$ and the constraint $g=0$.
2. Set $\nabla f=\lambda\nabla g$ and write all gradient equations.
3. Solve for $x,y,z$ in terms of $\lambda$ (or a single parameter $k$).
4. Substitute into $g=0$ to find $\lambda$.
5. Compute $f$ at each critical point; the smallest is the minimum.
6. Verify geometrically using the point-to-plane formula $d=|ax_0+by_0+cz_0-d|/\sqrt{a^2+b^2+c^2}$ as a check.

Worked Example(s)

2013 Paper 1, 2013-P1-Q3a (20 marks)

Find the shortest distance between the line $y=10-2x$ and the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$ by Lagrange multipliers.

The distance from $(x,y)$ to the line $2x+y-10=0$ is $|2x+y-10|/\sqrt5$. Minimise the squared distance $f=(2x+y-10)^2$ subject to $g=x^2/4+y^2/9-1=0$.

Let $k=2x+y-10$. The conditions $\nabla f=\lambda\nabla g$ give $4k=\lambda x/2$ and $2k=2\lambda y/9$, i.e. $\lambda x=8k,\qquad \lambda y=9k.$

The line does not meet the ellipse (discriminant of $25x^2-160x+364=0$ is $-10\,800<0$), so $k\ne0$, $\lambda\ne0$. Solve: $x=8k/\lambda$, $y=9k/\lambda$. Substitute into $g=0$: $\frac{64k^2}{4\lambda^2}+\frac{81k^2}{9\lambda^2}=\frac{25k^2}{\lambda^2}=1\;\Longrightarrow\;\lambda=\pm5k.$

Two critical points:

• $\lambda=5k$: $x=8/5$, $y=9/5$; then $k=16/5+9/5-10=-5$; distance $=5/\sqrt5=\sqrt5$.
• $\lambda=-5k$: $x=-8/5$, $y=-9/5$; then $k=-15$; distance $=15/\sqrt5=3\sqrt5$.

$\boxed{\;\text{Shortest distance }=\sqrt5,\;\text{at }\Bigl(\tfrac{8}{5},\tfrac{9}{5}\Bigr)\text{ on the ellipse.}\;}$

2020 Paper 1, 2020-P1-Q4c (20 marks)

Find the extreme value of $u=x^2+y^2+z^2$ subject to $2x+3y+5z=30$ by Lagrange’s method.

Set $\nabla u=\lambda\nabla g$ with $g=2x+3y+5z-30$: $2x=2\lambda,\quad 2y=3\lambda,\quad 2z=5\lambda\;\Longrightarrow\;x=\lambda,\quad y=\tfrac{3\lambda}{2},\quad z=\tfrac{5\lambda}{2}.$ Constraint: $2\lambda+\tfrac{9\lambda}{2}+\tfrac{25\lambda}{2}=19\lambda=30$, so $\lambda=30/19$.

$x=\frac{30}{19},\quad y=\frac{45}{19},\quad z=\frac{75}{19},\qquad u_{\min}=\frac{900+2025+5625}{361}=\frac{450}{19}.$

This is a minimum (the constraint is a plane; $u\to\infty$ as the point recedes; no maximum exists). Geometric check: perpendicular distance from origin to $2x+3y+5z=30$ is $30/\sqrt{38}$, giving $d^2=900/38=450/19$ ✓.

$\boxed{\;u_{\min}=\frac{450}{19},\;\text{at }\Bigl(\tfrac{30}{19},\tfrac{45}{19},\tfrac{75}{19}\Bigr).\;}$

Common Traps

• Minimise $(2x+y-10)^2$, not the distance itself; the critical points are identical and the algebra is cleaner. Recover the true distance by dividing by $\sqrt{a^2+b^2}$ at the end.
• In the 2013 problem, verify the line and ellipse do not intersect (discriminant $<0$); if they did, the minimum distance would be zero and Lagrange would be unnecessary.
• In the 2020 problem, state explicitly that this is a minimum only — there is no maximum on a plane.
• Arithmetic: $2\lambda+\tfrac{9\lambda}{2}+\tfrac{25\lambda}{2}=19\lambda$, not $17\lambda$ — a common slip.

applied-constrained-optimization (2 question(s); 2015, 2022)

Recognition Cues

• A word problem involving a geometric shape (cone, box, wire) with a fixed quantity (volume, perimeter, length).
• “Find the dimensions for which the [canvas / area / cost] is least, given that the [volume / length] is fixed.”
• One constraint (the fixed quantity), one objective (the quantity to optimise).

Solution Template

1. Introduce variables for the dimensions and write $f$ (objective) and $g=0$ (constraint) explicitly.
2. Work with $f^2$ or a convenient power if it avoids square roots.
3. Set $\nabla f=\lambda\nabla g$ and solve for the ratio of variables.
4. Use the constraint to recover actual values or the requested ratio.
5. State the nature (min or max) by convexity, second derivative, or boundary argument.

Worked Example(s)

2015 Paper 1, 2015-P1-Q2b (13 marks)

A conical tent is of given capacity. Find the ratio of height to base radius for the least canvas.

Let $r$ = base radius, $h$ = height. Volume constraint: $V=\frac{1}{3}\pi r^2h$ (fixed). Canvas = lateral surface: $S=\pi r\sqrt{r^2+h^2}$ (minimise). Work with $S^2$ to eliminate the square root: $S^2=\pi^2r^4+\frac{9V^2}{r^2}\qquad\text{(after substituting }h=3V/(\pi r^2)\text{)}.$ Set $\tfrac{d}{dr}S^2=0$: $4\pi^2r^3=18V^2/r^3\Rightarrow r^6=9V^2/(2\pi^2)$.

Then $h^2/r^2=9V^2/(\pi^2r^6)=2$, so $\boxed{\;\frac{h}{r}=\sqrt{2}.\;}$ $S^2$ is convex ($S'''>0$ for all $r>0$), so this is the global minimum.

2022 Paper 1, 2022-P1-Q2b (15 marks)

A wire of length $l$ is cut into two parts bent into a square and a circle. Use Lagrange’s method to find the least sum of areas.

Let the square use length $x$ and the circle use length $y$, with $x+y=l$. Areas: $A_1=x^2/16$ (side $=x/4$), $A_2=y^2/(4\pi)$ (radius $=y/(2\pi)$). Minimise $f=x^2/16+y^2/(4\pi)$ subject to $g=x+y-l=0$.

$\nabla f=\lambda\nabla g$: $\frac{x}{8}=\lambda,\quad\frac{y}{2\pi}=\lambda\;\Longrightarrow\;y=\frac{\pi x}{4}.$ Constraint: $x(1+\pi/4)=l\Rightarrow x=4l/(4+\pi)$, $y=\pi l/(4+\pi)$.

$f_{\min}=\frac{l^2}{(4+\pi)^2}+\frac{\pi l^2}{4(4+\pi)^2}=\frac{l^2}{(4+\pi)^2}\cdot\frac{4+\pi}{4}=\boxed{\;\frac{l^2}{4(4+\pi)}.\;}$

$f$ is strictly convex (sum of quadratics), confirming this is the global minimum.

Common Traps

• “Canvas” for a conical tent means lateral surface only ($\pi r\ell$, no floor). Using total surface area is a common error.
• In the wire problem, the square side is $x/4$ (perimeter $=4\times$side$=x$); the circle radius is $y/(2\pi)$. Errors here cascade into wrong areas.
• The 2015 problem can be solved by direct substitution without invoking Lagrange; both approaches are acceptable unless the question specifies.
• The 2022 problem explicitly requires Lagrange — show the $\nabla f=\lambda\nabla g$ setup even if you verify the answer by substitution.

symmetric-constrained (2 question(s); 2019, 2025)

Recognition Cues

• The objective is a symmetric function of all variables: $x^2y^2z^2$, $xyz$, or the volume $8xyz$ of an inscribed box.
• The constraint is symmetric: $x^2+y^2+z^2=c^2$ or an ellipsoid $x^2/a^2+y^2/b^2+z^2/c^2=1$.
• Often stated with $x,y,z>0$ (first-octant restriction).

Solution Template

1. Set $\nabla f=\lambda\nabla g$ and write the three gradient equations.
2. Multiply each equation by the corresponding variable; observe all three products are equal.
3. Conclude $x^2/a^2=y^2/b^2=z^2/c^2$ (or $x=y=z$ on a sphere).
4. Substitute into the constraint to find the common value.
5. Confirm the nature: boundary (any variable $=0$) gives $f=0$; the interior critical point is the maximum.

Worked Example(s)

2019 Paper 2, 2019-P2-Q4a (15 marks)

Find the maximum value of $f=x^2y^2z^2$ subject to $x^2+y^2+z^2=c^2$, $x,y,z>0$.

$\nabla f=\lambda\nabla g$: $2xy^2z^2=2\lambda x$, etc. Since $x,y,z>0$, divide by $2x,2y,2z$: $y^2z^2=\lambda,\quad x^2z^2=\lambda,\quad x^2y^2=\lambda\;\Longrightarrow\;x=y=z.$ Constraint: $3x^2=c^2\Rightarrow x=c/\sqrt3$. Maximum: $f_{\max}=\left(\frac{c^2}{3}\right)^3=\frac{c^6}{27}.$ Boundary: any variable $=0$ gives $f=0<c^6/27$, confirming this is the maximum. $\boxed{\;f_{\max}=\frac{c^6}{27},\;\text{at }x=y=z=\frac{c}{\sqrt3}.\;}$

AM–GM check: with $u=x^2,v=y^2,w=z^2$: $\sqrt[3]{uvw}\le(u+v+w)/3=c^2/3$, so $uvw\le c^6/27$, equality at $u=v=w$ ✓.

2025 Paper 2, 2025-P2-Q3b (20 marks)

Show that the volume of the greatest rectangular parallelepiped inscribed in $x^2/a^2+y^2/b^2+z^2/c^2=1$ is $8abc/(3\sqrt3)$.

The box centred at the origin with first-octant vertex $(x,y,z)$ on the ellipsoid has volume $V=8xyz$ (8 octants). Maximise $V$ subject to $g=x^2/a^2+y^2/b^2+z^2/c^2-1=0$.

$\nabla V=\lambda\nabla g$: $8yz=2\lambda x/a^2$, etc. Multiply each by $x,y,z$ respectively — all three give $8xyz=2\lambda(\ldots)$, forcing $\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}.$ Call this $t$; constraint gives $3t=1\Rightarrow t=1/3$, so $x=a/\sqrt3,\,y=b/\sqrt3,\,z=c/\sqrt3$.

$\boxed{\;V_{\max}=8\cdot\frac{abc}{3\sqrt3}=\frac{8\sqrt3\,abc}{9}.\;}$

Common Traps

• Factor of 8: the inscribed box has 8 vertices; volume $=8xyz$ where $(x,y,z)$ is the first-octant vertex. Forgetting this gives $abc/(3\sqrt3)$ — one-eighth of the correct answer.
• On the ellipsoid: equal ratios $x^2/a^2=y^2/b^2=z^2/c^2$, not equal values. Only when $a=b=c$ do we get $x=y=z$.
• AM–GM is a valid two-line alternative for the 2019 sphere problem and can serve as a check.
• State the boundary argument explicitly: $f=0$ at boundary, positive at the critical point — confirms maximum.

quadratic-form-extrema (2 question(s); 2014, 2016)

Recognition Cues

• Extremise $x^2+y^2+z^2$ subject to two constraints: an ellipsoid ($ax^2+by^2+cz^2=1$) and a plane ($lx+my+nz=0$).
• Often phrased “find the maximum and minimum values of $x^2+y^2+z^2$…” with a geometric interpretation requested.
• The presence of both a quadric and a homogeneous plane constraint is the signal.

Solution Template

1. Introduce two multipliers: $\nabla(r^2)=\lambda\nabla g+\mu\nabla h$.
2. Write the three gradient equations.
3. Key trick: multiply each equation by its variable and sum; using both constraints gives $\lambda=r^2$.
4. Express $x,y,z$ in terms of $\mu$ and $r^2$; apply the plane constraint to get the secular equation: $\frac{l^2}{1-ar^2}+\frac{m^2}{1-br^2}+\frac{n^2}{1-cr^2}=0.$
5. Clear denominators; solve the resulting quadratic in $r^2$ for the two extreme values.
6. Geometric interpretation: the constraint intersection is a central elliptical section of the ellipsoid; $r_1,r_2$ are its semi-axes.

Worked Example(s)

2014 Paper 1, 2014-P1-Q3b (20 marks)

Find max/min of $x^2+y^2+z^2$ subject to $ax^2+by^2+cz^2=1$ and $lx+my+nz=0$. Interpret geometrically.

With multipliers $\lambda,\mu$, the gradient equations are $2x(1-a\lambda)=\mu l$, and analogously for $y,z$. Multiply by $x,y,z$ and sum, then substitute both constraints: $(x^2+y^2+z^2)-\lambda\underbrace{(ax^2+by^2+cz^2)}_{=1}=\frac{\mu}{2}\underbrace{(lx+my+nz)}_{=0}\;\Longrightarrow\;\lambda=r^2.$

Solving: $x=\mu l/[2(1-ar^2)]$, etc. Substitute into the plane condition $lx+my+nz=0$: $\boxed{\;\frac{l^2}{1-ar^2}+\frac{m^2}{1-br^2}+\frac{n^2}{1-cr^2}=0.\;}$ This quadratic in $r^2$ yields the two squared extreme distances $r_1^2$ and $r_2^2$.

Geometric interpretation. The surface $ax^2+by^2+cz^2=1$ is an ellipsoid; the plane $lx+my+nz=0$ through its centre cuts it in a central ellipse. The extreme values $r_1$ and $r_2$ are the semi-major and semi-minor axes of that ellipse.

2016 Paper 1, 2016-P1-Q3a (20 marks)

Find max/min of $x^2+y^2+z^2$ subject to $\dfrac{x^2}{4}+\dfrac{y^2}{5}+\dfrac{z^2}{25}=1$ and $x+y-z=0$.

Here $a_1=4,a_2=5,a_3=25$ and $(l,m,n)=(1,1,-1)$. The secular equation is: $\frac{4}{4-r^2}+\frac{5}{5-r^2}+\frac{25}{25-r^2}=0.$ Clearing denominators gives $17r^4-245r^2+750=0$. Discriminant $=245^2-4(17)(750)=9025=95^2$: $r^2=\frac{245\pm95}{34}\;\Longrightarrow\;r^2=10\;\text{or}\;r^2=\frac{75}{17}.$

$\boxed{\;\max(x^2+y^2+z^2)=10,\qquad\min(x^2+y^2+z^2)=\frac{75}{17}.\;}$

Common Traps

• Two constraints require two multipliers. The secular equation is the result of eliminating both $\lambda$ and $\mu$; using only one multiplier is a fundamental error.
• The secular equation has denominators $(1-ar^2)$ or equivalently $(a_i-r^2)$ depending on normalisation — keep the form consistent when clearing denominators.
• In the 2016 example, the quadratic is $17r^4-245r^2+750=0$. Verify: discriminant $=9025=95^2$ (clean square root — if yours isn’t clean, check the algebra).
• The geometric interpretation (“semi-axes of central ellipse”) is required in the 2014 question — state it explicitly and name the ellipsoid.

Marks-Aware Writing

13-mark questions (2015): Set up in 2–3 lines, minimise $S^2$ via a single-variable derivative, state the ratio, confirm minimum by convexity. No geometry required.

15-mark questions (2019, 2022): Full Lagrange setup — write $L$, state $\nabla f=\lambda\nabla g$, solve the system, apply the constraint, compute the extremum, confirm nature. Box the answer.

20-mark questions (2013, 2014, 2016, 2020, 2025): Introduce notation (1–2 lines), write all gradient equations clearly, show the key algebraic step (identification of $\lambda$, or the two-case split), apply the constraint, compute all critical values, state which is max/min and why. For 2014: include the geometric interpretation paragraph. For the secular-equation problems (2014, 2016): display the quadratic explicitly and show the discriminant.

Practice Set