4.3 Lagrange Multipliers

Most optimization in practice comes with a leash: maximize $f$ , subject to $g = c$ . The leash changes everything. A constrained maximum is almost never at a critical point of $f$ itself — our specimen $f(x, y) = xy$ has only a saddle, no maximum at all — yet on the constraint circle $x^2 + y^2 = 4$ the problem has clean, finite answers. The question is where to look, and the figure answers it twice over: once on the map, once along a ribbon.

A translucent red-and-blue saddle surface, z equals x times y, with seventeen sky-blue contour lines on the plane z equals 0: hyperbolas, plus a crossed pair of straight lines at level zero. A flat amber circle of radius 2 lies in that plane, and a red curve traces the value of x times y above and below each point of the circle, rising to two crests of height 2 and dipping to two troughs of depth 2. Two red marker points at the crests are hidden in this view; they can be shown by opening the figure in XYZ 3D. — The translucent red-and-blue saddle $z = xy$ , its seventeen sky-blue level curves on the plane $z = 0$ (hyperbolas, plus a crossed pair of lines at level $0$ ), a flat amber circle of radius $2$ in that same plane — the constraint $x^2 + y^2 = 4$ — and a red curve, the *ribbon*, that follows the circle while rising and falling with the surface: $f$ evaluated along the constraint, the only part of the surface the problem is allowed to use. Two marker points at the constrained maxima are hidden in this view; step 4 shows them in the XYZ 3D editor.

Explore

Orbit to a low side view and follow the red ribbon once around. How many crests and how many troughs does it have? Note roughly where the crests sit over the amber circle.
A crest is where $f$ , walking along the constraint, stops increasing and starts decreasing — it stalls. What is the ribbon's slope, measured along the walk, at a stall?
The ribbon's height over the circle is $xy$ . Using symmetry — the product is largest where $x$ and $y$ are equal and share a sign — predict the exact coordinates of the two crests and the value of $f$ there. Commit before revealing.
Press Open in XYZ 3D ↗ under the figure, and in the editor's object list turn on the two hidden Point rows with their eye toggles. They sit at the crests; zoom in and read their positions against the grid. Do they match your $\pm(\sqrt{2}, \sqrt{2})$ , $f = 2$ ?
Now the map view, still in the editor: step the Contours row's z offset box (steps of $0.5$ ) from $0$ up to $2$ . The whole family of level curves rides up the saddle as one flat sheet. At crest height, the raised curve $xy = 2$ meets the red ribbon at exactly the two marked points — does it cross the ribbon there, or kiss it tangentially? Then set the z offset back to $0$ and look straight down: on the floor, the same curve $xy = 2$ meets the amber circle directly below the markers, in the same way.
Set the z offset to $1$ : the raised curve $xy = 1$ meets the ribbon at four points, crossing it each time. Over those four points, is the ribbon at a crest, or still rising or falling? Back at z offset $0$ , the same curve makes four honest crossings of the amber circle. Say the connection out loud: crossing a level curve means the walk is still changing $f$ . (The 17 levels put drawn curves exactly at the integer values $-8, \ldots, 8$ , so both curves in this step and the last are actually drawn; other level counts can hide them.)

From kiss to equations

At a tangency the constraint curve and the level curve share a tangent line, so their normals point along one line. Those normals are gradients — $\nabla f$ for the level curve, $\nabla g$ for the constraint — so at a constrained extremum

$\nabla f = \lambda\, \nabla g$

for some scalar $\lambda$ , the Lagrange multiplier, solved together with $g = c$ . The logic runs through step 6: where the level curve crosses the constraint, level curves of larger and smaller value lie on either side, so sliding along the constraint still changes $f$ — no extremum there. Only a kiss can stop the walk, and the ribbon says the same thing in its own language: kisses on the map are stalls on the ribbon.

Example 1 The full Lagrange computation on the circle

Find the maximum and minimum of $f(x, y) = xy$ on the circle $x^2 + y^2 = 4$ , with the multiplier at each.

Show solution

With $g = x^2 + y^2$ , the equations $\nabla f = \lambda \nabla g$ and the constraint read

$y = 2\lambda x, \qquad x = 2\lambda y, \qquad x^2 + y^2 = 4.$

Substituting the first into the second gives $x = 4\lambda^2 x$ . If $x = 0$ the first equation forces $y = 0$ , violating the constraint — so $\lambda^2 = \tfrac{1}{4}$ , i.e. $\lambda = \pm\tfrac{1}{2}$ . For $\lambda = \tfrac{1}{2}$ the equations give $y = x$ , and the constraint gives $2x^2 = 4$ : the points $(\sqrt{2}, \sqrt{2})$ and $(-\sqrt{2}, -\sqrt{2})$ , where $f = 2$ . For $\lambda = -\tfrac{1}{2}$ , $y = -x$ yields $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$ , where $f = -2$ . So the constrained maximum is $2$ and the minimum $-2$ . The ribbon confirms it in one variable: on the circle, $f(2\cos t, 2\sin t) = 4\cos t \sin t = 2\sin 2t$ , which crests at $2$ and bottoms at $-2$ , four stalls per lap — exactly the four kisses.

\frac{1}{2},\; -\frac{1}{2}

The multiplier equation, solved ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The pivot of the example: eliminating $x$ and $y$ from the Lagrange equations leaves $4\lambda^2 = 1$ (written here with $L$ for $\lambda$ ), whose two roots $\lambda = \pm\tfrac{1}{2}$ hand back the lines $y = x$ and $y = -x$ and, with the constraint, all four kiss points.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.