Multivariable Calculus on the Putnam

The Putnam almost never wants a double integral computed by brute force. The integrand is chosen so that a symmetry of the region turns it into something trivial: $\iint_{[0,1]^2}\frac{x}{x + y}\,dA$ averages with its mirror to $\iint\frac12 = \frac12$. Whenever the region is a square, a disk, or a simplex, list its symmetries FIRST and apply each to the integrand.

A Jacobian is not a nuisance factor — it is the reason the trick works. The polar $r$ is what makes $\int e^{-x^2}$ computable; the $\tfrac12$ from $u = x + y$, $v = x - y$ is what turns $\iint_{[0,1]^2}|x - y|\,dA$ into a one-variable integral. The other classic: $\iint$ over the triangle $x, y \ge 0$, $x + y \le 1$ of $f(x + y)$ equals $\int_0^1 t f(t)\,dt$, because the level sets $x + y = t$ have length proportional to $t$.

The Putnam version usually has symmetric $f$ and $g$, so the candidates are points with several equal coordinates — but the maximum is not always at the fully symmetric point. Compute $f$ at every candidate. For a sum $\sum x_i^2$ with $\sum x_i$ fixed, the extremes are "all equal" and "one variable takes everything", and Lagrange tells you which is which.

These appear on the Putnam as identities to recognize, not theorems to apply mechanically. The area formula turns "area of a polygon with vertices…" into the shoelace formula; the flux fact makes a hideous surface integral equal $4\pi$ or $0$ depending only on whether the surface encloses the origin.

The Jacobian determinant measures how the map from $(u, v)$ to $(x, y)$ stretches area (see "Why It Works" below): a tiny $du\times dv$ rectangle becomes a parallelogram of area $|J|\,du\,dv$. For a linear substitution $J$ is a constant, so it simply scales the answer — which is why forgetting it gives an answer off by a fixed factor. Before computing, list what the substitution does to each edge of the region; a substitution that simplifies the integrand but scrambles the region has not helped.

Step 1 — substitute $x = au$, $y = bv$. The region becomes the unit disk $u^2 + v^2 \le 1$.

Step 2 — the Jacobian is $\det\begin{pmatrix}a&0\\0&b\end{pmatrix} = ab$.

Step 3 — area $= \iint_{u^2 + v^2 \le 1}ab\,du\,dv = ab\cdot\pi = \pi ab$. For $a = b$ it is the circle's $\pi a^2$.

How to find it. A linear map that sends a hard region to an easy one multiplies every area by $|\det|$: that single number is all the integral needs.

Step 1 — the exponent is $\tfrac uv$ with $u = y - x$, $v = y + x$. Inverting, $x = \tfrac{v - u}{2}$, $y = \tfrac{u + v}{2}$, and $|J| = \left|\det\begin{pmatrix}-\tfrac12 & \tfrac12\\ \tfrac12 & \tfrac12\end{pmatrix}\right| = \tfrac12$.

Step 2 — the region: $x \ge 0$ is $u \le v$, $y \ge 0$ is $u \ge -v$, and $x + y \le 1$ is $v \le 1$. So $T$ becomes $0 \le v \le 1$, $-v \le u \le v$.

Step 3 — integrate in $u$ first: $\int_{-v}^{v}e^{u/v}\,du = v\left(e - e^{-1}\right)$.

Step 4 — then $\tfrac12\int_0^1 v\left(e - e^{-1}\right)dv = \tfrac{e - e^{-1}}{4} = \tfrac{\sinh 1}{2} \approx 0.588$.

How to find it. The integrand is constant on rays through the origin (dashed), so one new variable should measure the ray ($\tfrac uv$) and the other the distance along it ($v$). Integrating over $u$ at fixed $v$ sweeps across the rays, where the exponential integrates in closed form.

Step 1 — complete the square: $x^2 + xy + y^2 = \left(x + \tfrac y2\right)^2 + \tfrac34y^2$.

Step 2 — substitute $u = x + \tfrac y2$, $w = \tfrac{\sqrt3}{2}y$. Then $\tfrac{\partial(u, w)}{\partial(x, y)} = \det\begin{pmatrix}1 & \tfrac12\\ 0 & \tfrac{\sqrt3}{2}\end{pmatrix} = \tfrac{\sqrt3}{2}$, so $dx\,dy = \tfrac{2}{\sqrt3}\,du\,dw$, and the plane maps onto the plane.

Step 3 — $\iint e^{-(u^2 + w^2)}\,du\,dw = \pi$ (polar coordinates, or the square of the Gaussian integral).

Step 4 — the answer is $\tfrac{2\pi}{\sqrt3} \approx 3.628$. In general $\iint e^{-\mathbf x^{\mathsf T}A\mathbf x}\,dA = \tfrac{\pi}{\sqrt{\det A}}$ for positive definite $A$; here $A = \begin{pmatrix}1 & \tfrac12\\ \tfrac12 & 1\end{pmatrix}$, $\det A = \tfrac34$.

How to find it. A quadratic form with a cross term is a circle in disguise. Completing the square (or diagonalizing $A$: eigenvalues $\tfrac32$ and $\tfrac12$ along $y = \pm x$) turns it into a sum of squares, and the Jacobian records the determinant.

If $\nabla f$ had a component along the constraint set at a point $p$, moving from $p$ along the set in that direction would increase $f$ and moving the other way would decrease it, so $p$ would not be an extremum. Hence $\nabla f$ is perpendicular to the set, and so is $\nabla g$; in the plane, perpendicular to the same curve means parallel. Geometrically, the level curve of $f$ through an extremum is TANGENT to the constraint. The method produces candidates, not answers: it says nothing about points where $\nabla g = 0$, and it does not say which candidate is the maximum — on a compact set the extreme value theorem guarantees both extremes exist, and comparing the values finishes the problem.

Step 1 — the ellipse is compact, so both extremes exist, and $\nabla g = \left(\tfrac x4, y\right) \ne 0$ on it.

Step 2 — $\nabla(xy) = (y, x) = \lambda\left(\tfrac x4, y\right)$ gives $y = \tfrac{\lambda x}{4}$ and $x = \lambda y$. If $y = 0$ then $x = 0$, not on the ellipse; so $y \ne 0$ and $y = \tfrac{\lambda^2}{4}y$, i.e. $\lambda = \pm2$ and $x = \pm2y$.

Step 3 — on the ellipse, $\tfrac{4y^2}{8} + \tfrac{y^2}{2} = y^2 = 1$, so the candidates are $(\pm2, \pm1)$ with all four sign choices.

Step 4 — $xy = 2$ at $(2, 1)$ and $(-2, -1)$, and $-2$ at the other two. Maximum $2$, minimum $-2$. (Check by AM–GM: $1 = \tfrac{x^2}{8} + \tfrac{y^2}{2} \ge 2\sqrt{\tfrac{x^2y^2}{16}} = \tfrac{|xy|}{2}$.)

How to find it. Picture the level curves $xy = k$ growing until they last touch the ellipse; the touching points are where Lagrange's condition holds, and symmetry ($x \to -x$, $y \to -y$) predicts four of them.

Step 1 — the disk is compact and $f$ continuous: both extremes exist.

Step 2 — interior: $\nabla f = (2xy, x^2) = 0$ forces $x = 0$, where $f = 0$.

Step 3 — boundary: on $x^2 + y^2 = 3$, $f = (3 - y^2)y = h(y)$ for $y \in [-\sqrt3, \sqrt3]$. $h'(y) = 3 - 3y^2 = 0$ at $y = \pm1$, with $h(\pm1) = \pm2$; at the ends $h(\pm\sqrt3) = 0$.

Step 4 — compare all candidates: $0$ (interior), $\pm2$, $0$. The maximum is $2$ at $(\pm\sqrt2, 1)$ and the minimum is $-2$ at $(\pm\sqrt2, -1)$.

How to find it. On a region the boundary is a separate problem one dimension lower; parametrizing it (here by $y$) often beats Lagrange. The interior critical points of a polynomial like $x^2y$ are usually degenerate and rarely win.

Step 1 — find the range of one variable. Given $z$, the other two have $x + y = 5 - z$ and $xy = 8 - z(x + y) = 8 - 5z + z^2$. They are real iff the discriminant of $t^2 - (5 - z)t + (8 - 5z + z^2)$ is $\ge 0$: $(5 - z)^2 - 4(8 - 5z + z^2) = -3z^2 + 10z - 7 = -(3z - 7)(z - 1) \ge 0$, i.e. $1 \le z \le \tfrac73$.

Step 2 — the target in terms of $z$: $xyz = z(8 - 5z + z^2) = z^3 - 5z^2 + 8z = p(z)$.

Step 3 — $p'(z) = 3z^2 - 10z + 8 = (3z - 4)(z - 2)$, so the candidates on $[1, \tfrac73]$ are $z = 1, \tfrac43, 2, \tfrac73$, with $p = 4, \tfrac{112}{27}, 4, \tfrac{112}{27}$.

Step 4 — the maximum is $\tfrac{112}{27}$ (at $(\tfrac43, \tfrac43, \tfrac73)$ and permutations) and the minimum is $4$ (at $(2, 2, 1)$ and permutations). In both, two variables are equal — as Lagrange with two constraints predicts — but which equal pair wins is decided only by computing: the configuration with the larger lone variable wins the maximum.

How to find it. With the sum and the pairwise sum fixed, the three numbers are the roots of $t^3 - 5t^2 + 8t - p$, and $p = xyz$ is the only free parameter; the cubic has three real roots exactly on an interval of $p$, whose ends are where two roots merge. Reducing to one variable makes that interval visible.