Multivariable Calculus: Lagrange Multipliers

Podcast

Play an AI-generated podcast conversation about this lesson

Questions and Answers

What principle underlies the method of Lagrange multipliers when finding extreme values of a function $f(x, y)$ subject to a constraint $g(x, y) = k$?

The extreme values are unrelated to the gradients of $f$ and $g$, and depend only on the value of $k$.
The extreme values occur where level curves of $f$ cross level curves of $g$ at right angles.
At the extreme values, the gradient of $f$ is perpendicular to the gradient of $g$.
At the extreme values, the level curves of $f$ and $g$ are tangent to each other. (correct)

In the context of Lagrange multipliers, what does the scalar $\lambda$ (lambda) represent?

The rate of change of the constraint function $g(x, y, z)$.
The minimum value of the function $f(x, y, z)$.
The maximum value of the function $f(x, y, z)$.
A scaling factor relating the gradients of $f$ and $g$ at the point of optimization. (correct)

When using Lagrange multipliers to find the extreme values of $f(x, y, z)$ subject to $g(x, y, z) = k$, why is it necessary to ensure that $\nabla g \neq 0$?

Because the method is undefined when the gradient of the constraint function is zero, indicating a critical point of $g$ itself. (correct)
Because the extreme values of $f(x, y, z)$ will always occur where the gradient of $g$ is non-zero.
To ensure that $f(x, y, z)$ also has a non-zero gradient.
To simplify the calculations in the Lagrange multiplier equations.

If applying Lagrange multipliers yields multiple points $(x, y, z)$ that satisfy the Lagrange conditions, how do you determine which point corresponds to the maximum value of $f(x, y, z)$?

Evaluate $f(x, y, z)$ at each point and select the point yielding the largest value. (D) Signup and view all the answers

Consider the problem of finding the extreme values of $f(x,y) = xy^2$ subject to the constraint $x^2 + y^2 = 3$. Which system of equations needs to be solved using Lagrange multipliers?

$y^2 = 2\lambda x, 2xy = 2\lambda y, x^2 + y^2 = 3$ (B) Signup and view all the answers

What is a key difference in applying Lagrange multipliers to functions of two variables versus functions of three variables?

The fundamental principle remains the same (gradients are parallel), but the number of equations to solve increases with each additional variable. (A) Signup and view all the answers

Suppose you are using Lagrange multipliers to find the minimum value of a function $f(x, y)$ subject to constraint $g(x, y) = k$. After solving the Lagrange equations, you find only one possible point $(x_0, y_0)$. What can you conclude?

If the problem is well-posed and the function and constraint satisfy the conditions for Lagrange multipliers, $(x_0, y_0)$ is either the minimum, or there is no minimum. (D) Signup and view all the answers

Which of the following is a correct setup for finding the extreme values of $f(x, y, z) = x + y + z$ subject to the constraint $x^2 + y^2 + z^2 = 1$ using Lagrange multipliers?

$1 = 2\lambda x, 1 = 2\lambda y, 1 = 2\lambda z, x^2 + y^2 + z^2 = 1$ (D) Signup and view all the answers

Why do we need to consider the case when the level curves of f(x, y) and g(x, y) have a common tangent line, when using Lagrange multipliers?

Because, at the point of tangency, the gradients of $f(x, y)$ and $g(x, y)$ are parallel, indicating a potential maximum or minimum. (D) Signup and view all the answers

Suppose you want to find the point on the plane $x + y + z = 1$ that is closest to the origin using Lagrange multipliers. Which function would you minimize subject to the given constraint?

$f(x, y, z) = x^2 + y^2 + z^2$ (B) Signup and view all the answers

Flashcards

Lagrange Multipliers

A method for finding the maximum or minimum values of a function subject to constraints.

Lagrange Multipliers Method (2 Variables)

Find where ▽f(x, y) = λ▽g(x, y) and g(x, y) = k. Then evaluate f at all points.

Lagrange Multipliers Method (3 Variables)

Find all x, y, z, and λ such that ▽f(x, y, z) = λ▽g(x, y, z) and g(x, y, z) = k. Evaluate f at these points.

Geometric Interpretation

Extreme values occur where level curves of f(x, y) are tangent to the constraint curve g(x, y) = k.

Signup and view all the flashcards

System of Equations

Solving this system helps identify candidate points for maxima and minima under the constraint.

Signup and view all the flashcards

Study Notes

Math 2110Q Multivariable Calculus (Spring 2025)
1. 8 Lagrange Multipliers

Learning Objectives

Understand the method of Lagrange Multipliers.
Use Lagrange Multipliers to find the max/min of function over a constrained domain.

Function of Two Variables

To find the extreme values of z = f(x, y), use a constraint of the form g(x, y) = k.
View g(x, y) = k as level curves of surface z = g(x, y) with z = k and consider level curves of f(x, y).
Maximizing f(x, y) subject to g(x, y) = k involves finding the largest value of c such that the level curve f(x, y) = c intersects g(x, y) = k.
These two curves will just touch each other. That is when they have a common tangent line.
Normal vectors at the intersection point (x0, y0) are parallel, so gradient vectors at (x0, y0) are parallel. Results in ▽f(x0, y0) = λ▽g(x0, y0) for some scalar λ.

Functions of Three Variables

The Method of Lagrange Multipliers can be used to find the maximum/minimum values of f(x, y, z) subject to the constraint g(x, y, z) = k. Assuming that the extreme values exist and ▽g ≠ 0 on the surface g(x, y, z) = k.
Find all values of x, y, z, and λ such that:
- ▽f(x, y, z) = λ▽g(x, y, z)
- g(x, y, z) = k
Evaluate f at all the points (x, y, z) from the previous step. The largest value is the maximum of f; the smallest is the minimum value of f.

Examples

Find the extreme values of the function f(x, y) = xy² on the circle x² + y² = 3.
Find the extreme values of f(x, y) = x² + y² subject to xy = 1.
When we obtain one value by this method, this means that there is either a maximum with no minimum or vice versa.

Studying That Suits You

Use AI to generate personalized quizzes and flashcards to suit your learning preferences.