Lagrange Multipliers and Extrema

Lagrange Multipliers and Extrema

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

Created by

Mia Campbell

FREE Resource

The video tutorial explains how to use Lagrange multipliers to find the absolute extrema of a function subject to a constraint. It involves setting up and solving a system of equations using partial derivatives. The tutorial demonstrates solving for x and y, finding extrema points, and evaluating function values to determine the absolute maximum and minimum. Finally, it verifies the results graphically.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the constraint equation used in the problem?

x^2 + y^2 = 0

x^2 - y^2 = 0

x^2 - y^2 = 81

x^2 + y^2 = 81

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the form of the constraint equation g(x, y) used in Lagrange multipliers?

g(x, y) = x^2 - y^2 - 81

g(x, y) = x^2 - y^2 + 81

g(x, y) = x^2 + y^2 - 81

g(x, y) = x^2 + y^2 + 81

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the partial derivative of f with respect to x?

-2x + 5

2x + 5

-2y + 5

2y + 5

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you solve for x in the first equation?

Add 2x to both sides and divide by lambda

Subtract 2x from both sides and multiply by lambda

Add 2x to both sides and factor out 2x

Subtract 2x from both sides and factor out 2x

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the relationship between x and y found in the solution?

x = y/2

x = y

x = -y

x = 2y

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the possible values for x and y after solving the equations?

x = ±√2, y = ±√2

x = ±9, y = ±9

x = ±2, y = ±2

x = ±9/√2, y = ±9/√2

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the approximate function value when both x and y are positive?

-21.3604

148.6396

-148.6396

21.3604

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