Absolute extrema on a closed region and Lagrange multipliers

Last week’s second derivative test finds only local extrema. But in practice most questions ask “what is the highest and lowest value in this region?” or “what is the best value under this constraint?”. This week we answer these two questions. In the first (absolute extrema on a closed region) we scan the inside and the boundary of the region separately and compare the candidates. In the second (Lagrange multipliers) we find the candidates without solving the constraint, using the condition that the gradients are parallel. The exam has one big question from each; points are lost mostly by skipping a piece of the boundary or a solution case.

Golden rules:

1. On a closed and bounded region, a continuous function always has an absolute max and min (Extreme Value Theorem).
2. Candidates: interior critical points + critical points on the boundary + vertices.
3. For absolute extrema the $D$ test is not needed. Just compare the values.
4. Lagrange: $\nabla f = \lambda\nabla g$ and $g = k$. Before dividing, examine separately the factors that could be zero.

Absolute extrema: the closed region method

The region must be closed (it includes its boundary) and bounded (it does not go to infinity). A disk, a rectangle and a triangle are typical examples.

Method.

1) INTERIOR: solve f_x = 0, f_y = 0. Keep only the points that lie inside the region.
2) BOUNDARY: treat each edge separately.
     line segment → substitute y (or x), a single-variable function
       → set the derivative to 0 (roots inside the interval) + the two endpoints of the edge
     circle x²+y²=r² → substitute x² = r² - y², or x = r cos t, y = r sin t
3) TABLE: write all the candidates and the f values. Choose the largest and the smallest.

Example 1. $f(x,y) = x^2+y^2-2x-4y$, the triangle with vertices $(0,0)$, $(4,0)$, $(0,4)$.

Interior. $f_x = 2x-2 = 0$, $f_y = 2y-4 = 0$: $(1,2)$. $1+2\le4$, inside. $f(1,2) = 1+4-2-8 = -5$.

Edge 1, $y = 0$, $0\le x\le4$. $f = x^2-2x$, $f' = 2x-2 = 0$: $x = 1$, $f = -1$. Ends: $f(0,0) = 0$, $f(4,0) = 8$.

Edge 2, $x = 0$, $0\le y\le4$. $f = y^2-4y$, $y = 2$: $f = -4$. Ends: $0$ and $f(0,4) = 0$.

Edge 3, $x+y = 4$. $y = 4-x$: $f = x^2+(4-x)^2-2x-4(4-x) = 2x^2-6x$. $f' = 4x-6 = 0$: $x = 1.5$, the point $(1.5,2.5)$, $f = -4.5$. The ends are already vertices.

Absolute max $8$, at $(4,0)$. Absolute min $-5$, at $(1,2)$.

PITFALL · forgetting the vertices. When you reduce each edge to a single-variable function, the endpoints of that edge are candidates too. Since the endpoints are vertices, the easiest way is to compute the values at the vertices at the very start and write them in the table. In this example the maximum came out at a vertex, and no derivative zero gave it.

Lagrange multipliers

Problem. Find the largest and smallest values of $f(x,y)$ on the constraint $g(x,y) = k$ (a curve). In three variables the constraint is a surface: $g(x,y,z) = k$.

Theorem. At an extremum point of $f$ on the constraint (and if $\nabla g\neq\mathbf 0$) there is a number $\lambda$ such that

$\nabla f = \lambda\nabla g,\qquad g = k.$

In two variables these are three equations ($f_x = \lambda g_x$, $f_y = \lambda g_y$, $g = k$) in three unknowns ($x,y,\lambda$). In three variables, four equations in four unknowns.

Intuition. Imagine walking along the constraint curve and crossing the level curves of $f$. As long as you cross a level curve, $f$ is increasing or decreasing. At an extremum the constraint curve is tangent to the level curve of $f$. If two curves are tangent, their normals are parallel. The normals are the gradients: $\nabla f\parallel\nabla g$. $\lambda$ is the coefficient of parallelism.

Example 2. $f = xy$, constraint $x^2+y^2 = 8$.

∇f = <y, x>,  ∇g = <2x, 2y>
y = 2λx        (1)
x = 2λy        (2)
x² + y² = 8    (3)
Multiply (1) by x and (2) by y: xy = 2λx² and xy = 2λy²  →  λx² = λy²
If λ = 0, then from (1),(2) x = y = 0, which does not satisfy (3) → λ ≠ 0, so x² = y²
(3): 2x² = 8, x = ±2, y = ±2  → four points
f(2,2) = f(-2,-2) = 4,   f(2,-2) = f(-2,2) = -4

Max $4$, min $-4$. The constraint is a circle (closed and bounded), both values certainly exist.

Tactics for solving the system

TIP · zero cases. In an equation like $2xy = 2\lambda x$, before cancelling $x$ write the case $x = 0$ separately. In Q9 the $x = 0$ branch produces two candidates with the value $f = 0$. Even if they are neither the min nor the max, they must be in the list of candidates; skipping them leaves the solution incomplete.

Example 3 (three variables). The largest value of $f = x+y+z$ on the sphere $x^2+y^2+z^2 = 3$. $\langle1,1,1\rangle = \lambda\langle2x,2y,2z\rangle$, so $x = y = z = \frac{1}{2\lambda}$. Constraint: $3x^2 = 3$, $x = \pm1$. Max $f(1,1,1) = 3$, min $f(-1,-1,-1) = -3$.

Lagrange or substitution?

If you can easily pull a variable out of the constraint (a line, a plane), last week’s substitution method works too. Lagrange’s advantage: when the constraint is complicated (a circle, an ellipse, $xyz = 1$) it gives a symmetric system without taking roots and without splitting cases. In closed-region questions, if the boundary is a circle or an ellipse you can examine the boundary with Lagrange and the interior with $\nabla f = \mathbf 0$.

Common mistakes