Difficulty: Intermediate–Advanced | Prerequisites: Partial derivatives, gradients, single-constraint Lagrange multipliers (§14.8 first half)
Lagrange multipliers let you find the maximum or minimum of a function subject to one or more constraints. When you have two constraints, you introduce two multipliers and set the gradient of the objective equal to a linear combination of the constraint gradients. The algebra gets heavier, but the underlying logic is the same: constrained extrema occur where the gradients are linearly dependent.
Constraint
A condition that restricts the domain of the optimisation problem, expressed as g(x, y, z) = k. In simple terms, it is a rule that forces the solution to lie on a particular curve or surface.
Lagrange multiplier (lambda, mu)
A scalar introduced for each constraint. At a constrained extremum, the gradient of the objective function equals the sum of the multipliers times their respective constraint gradients. Think of it as measuring how much the constraint is "holding back" the objective function.
Gradient (nabla f)
The vector of all partial derivatives of f. It points in the direction of steepest increase. At a constrained extremum, the gradient of f must be a linear combination of the constraint gradients.
Hyperboloid of two sheets
A quadric surface that consists of two separate curved pieces. In standard form it looks like z squared over c squared minus x squared over a squared minus y squared over b squared equals 1. Think of two bowls facing away from each other.
Extremal values
The maximum and minimum values a function attains on a given region or subject to given constraints. These may occur in the interior (critical points), on the boundary, or at points satisfying the Lagrange conditions.
Feasible set
The collection of all points that satisfy every constraint simultaneously. When you have two constraints in three variables, the feasible set is typically a curve in space.
To optimise f(x, y, z) subject to g(x, y, z) = k, solve the system: nabla f = lambda times nabla g, together with g = k.
This gives one multiplier lambda and (in three variables) four equations in four unknowns.
Strategy: isolate ratios from the component equations to reduce unknowns, then substitute back into the constraint.
To optimise f(x, y, z) subject to g(x, y, z) = k1 and h(x, y, z) = k2, solve: nabla f = lambda times nabla g + mu times nabla h, together with both constraints.
This gives two multipliers and (in three variables) five equations in five unknowns: x, y, z, lambda, mu.
The feasible set is usually a curve (the intersection of two surfaces), so extrema along that curve are what you are finding.
Substitution before multipliers: When one constraint can be solved for a variable (e.g. y + z = 1 gives z = 1 - y), substitute it in to reduce the number of unknowns and constraints. Problem 3 on this set rewards this approach.
Symmetric structure: If the objective or constraints have symmetry, exploit it. Look for cases where two variables must be equal at the extremum.
Checking boundaries of a region: Problem 2 combines Lagrange multipliers on a boundary curve with checking interior critical points and edges of a triangular region. The absolute extrema of f on a closed bounded region come from comparing interior critical points, boundary critical points (via Lagrange), and corner values.
Step 1: Find interior critical points by setting nabla f = 0 inside the region.
Step 2: Parametrise or apply Lagrange multipliers on each boundary segment.
Step 3: Check all corner/intersection points.
Step 4: Compare all candidate values to identify the absolute max and min.
When the region is further restricted by a circle (the boundary of a disk), the feasible set becomes the arc of the circle inside the region. Parametrise using x = r cos(t), y = r sin(t) and restrict t appropriately.
The surface is defined implicitly by F(x, y, z) = 100.
Minimise x subject to the constraint F = 100. This is a standard single-constraint Lagrange problem.
Solve nabla x = lambda times nabla F, which gives 1 = lambda times partial F / partial x, and 0 = lambda times partial F / partial y, and 0 = lambda times partial F / partial z.
The equations 0 = lambda times (something) with lambda nonzero (since the first equation forces lambda to be nonzero) mean those partial derivatives must be zero. Use that to find relationships among x, y, z.
Single constraint
\nabla f = \lambda \nabla g \quad \text{subject to} \quad g(x,y,z) = kIn component form: f_x = lambda g_x, f_y = lambda g_y, f_z = lambda g_z, plus the constraint equation. Four equations, four unknowns.
Two constraints
\nabla f = \lambda \nabla g + \mu \nabla h \quad \text{subject to} \quad g = k_1, \; h = k_2In component form: f_x = lambda g_x + mu h_x, f_y = lambda g_y + mu h_y, f_z = lambda g_z + mu h_z, plus both constraint equations. Five equations, five unknowns.
Completing the square (useful for Problem 2)
f(x,y) = x^2 + y^2 - 2x - 4y = (x-1)^2 + (y-2)^2 - 5This reveals the unconstrained minimum is at (1, 2) with value -5. Knowing this helps you check whether the unconstrained minimum lies inside your feasible region.
"The multiplier lambda tells me the max or min value." It does not. Lambda measures sensitivity: how much the optimum changes per unit change in the constraint constant. The extreme value is f evaluated at the solution point.
"I need Lagrange multipliers for every boundary." Not always. If one constraint can be solved explicitly for a variable, substitution may be simpler and less error-prone than introducing another multiplier. Problem 3 on this set is a good example.
"Interior critical points cannot be extrema when there are constraints." They can, when the constraint region includes an interior. Problem 2 asks for the absolute extrema on a triangular region, so you must check interior critical points as well as boundary behaviour.
"Two constraints always give a unique solution." Two constraints in three variables typically reduce the feasible set to a curve, but that curve can yield multiple candidate points. You still need to evaluate f at each candidate and compare.
Lagrange multipliers are a staple of Calculus III exams. The method generalises to optimisation in any number of dimensions and underpins large parts of mathematical economics, physics, and engineering design.
Exams frequently test the two-constraint version with three variables, so practise setting up the five-equation system quickly.
A common exam format: "find the point on the surface closest to the origin" or "find the extreme values subject to two constraints." Both reduce to Lagrange problems.
Know when substitution simplifies the problem. If you spot that a constraint is linear in one variable, solve for it and reduce the problem before writing out the full gradient system.
Be ready to complete the square on quadratic objectives. It quickly reveals the unconstrained minimum, which tells you whether the minimum lies inside the feasible region or must occur on the boundary.
True or false: When optimising f subject to two constraints g = k1 and h = k2, you need exactly two Lagrange multipliers. True.
Fill in the blank: The gradient condition for two constraints is nabla f = ___ nabla g + ___ nabla h. Answer: lambda, mu (one multiplier per constraint).
True or false: If the unconstrained minimum of f lies inside the feasible region, then that point is also the constrained minimum. True (provided the region includes its interior, not just a boundary curve).
True or false: Lagrange multipliers can only find maxima, not minima. False. The method finds all constrained critical points; you determine which are maxima and which are minima by comparing values.
Fill in the blank: Two constraints in three variables typically restrict the feasible set to a ___. Answer: curve.
Q: A surface is defined by F(x, y, z) = c. How do you set up the Lagrange system to find the point on the surface with the smallest x-coordinate?
A: Minimise f(x, y, z) = x subject to F(x, y, z) = c. The gradient condition is (1, 0, 0) = lambda times (F_x, F_y, F_z). From the second and third components, lambda F_y = 0 and lambda F_z = 0. Since the first component forces lambda to be nonzero, you get F_y = 0 and F_z = 0. Solve these alongside F = c.
Q: Let f(x, y) = x squared + y squared minus 2x minus 4y and let A be the triangular region with x >= 0, 0 <= y <= 5, y >= x. Where does the absolute minimum of f on A occur?
A: Complete the square: f = (x - 1) squared + (y - 2) squared minus 5. The unconstrained minimum is at (1, 2) with value -5. Check that (1, 2) lies in A: x = 1 >= 0, y = 2, 0 <= 2 <= 5, and 2 >= 1. It does, so the absolute minimum on A is -5 at (1, 2). For the maximum, evaluate f at the corners and along each boundary edge.
Q: Optimise zx + 2xy subject to y + z = 1 and x squared + y squared = 1. What simplification should you try first?
A: The constraint y + z = 1 gives z = 1 - y. Substitute into the objective: (1 - y)x + 2xy = x - xy + 2xy = x + xy = x(1 + y). Now you have a single-constraint problem: optimise x(1 + y) subject to x squared + y squared = 1.
Q: Find the extreme values of f(x, y, z) = x squared + y squared + z squared subject to x minus y = 1 and y squared minus z squared = 1. How many equations and unknowns does the Lagrange system have?
A: Three gradient equations (one per variable) plus two constraint equations gives five equations. The unknowns are x, y, z, lambda, mu, also five. Solve the system, then evaluate f at each solution to identify the extrema.
This material builds directly on single-variable optimisation and single-constraint Lagrange multipliers from earlier in section 14.8. The gradient condition nabla f = lambda nabla g is the same idea as setting a derivative to zero in Calculus I, generalised to higher dimensions.
The boundary-checking technique in Problem 2 connects to the Extreme Value Theorem from section 14.7: a continuous function on a closed, bounded region attains its absolute maximum and minimum. In later sections on double integrals (15.1 and 15.2), you will describe these same regions as domains of integration.
In physics and engineering, constrained optimisation appears in thermodynamics (maximising entropy subject to energy constraints), structural engineering (minimising weight subject to strength requirements), and economics (maximising utility subject to a budget constraint).
Lagrange multipliers, constrained optimisation, constrained optimization, multiple constraints, two constraints, gradient condition, lambda mu, hyperboloid of two sheets, absolute max and min on a region, boundary of a region, extreme value theorem, completing the square, feasible set, MATH 215, Calculus III, section 14.8, University of Michigan