Lagrange multipliers problem

In summary: That's why the negative numbers are first in the lexicographic ordering.In summary, The problem involves finding the absolute maximum and minimum of the function f(x,y) = x^2-y^2 subject to the constraint x^2+y^2=289. The maximum is 289 and the minimum is -289, and the points are listed in lexicographic order. The correct answers are: max = 289 at points (-17, 0) and (17,0), and min = -289 at points (0,-17) and (0,17). The process involves taking the gradients and solving for lambda, which is either 1 or -1. When y = 0, lambda = 1
  • #1
glid02
54
0
Hey, I need help with a problem involving Lagrange multipliers...

Here is the question:
Find the absolute maximum and minimum of the function
f(x,y) = x^2-y^2 subject to the constraint x^2+y^2=289.

As usual, ignore unneeded answer blanks, and list points in lexicographic order.

I found that the max is 289 and the min is -289.

I took the gradients of each and came up with:
2x-2y = lambda2x+lambda2y
x=lambda(x)
-y=lambda(y)

Here's where I'm stuck, I know lambda = +/-1, but I can't come up with an equation to solve for x or y in terms of y or x. The best idea I can come up with is to solve for lambda from -y=lambda(y), which is just
-y/y and plug it into x=lambda(x), but that doesn't help any when solving for y.

I guess lambda = 1 when y = 0 and lambda = -1 when x = 0.

OK, I found the correct answers, but I could still use an explanation as to how to list them lexicographically.

The correct answer is:
max = 289 at points (-17, 0) and (17,0)
min = -289 at points (0,-17) and (0,17)

When y = 0, lambda = 1 so (lambda(x))^2+0=289
which is x^2=289, x = 17

When x = 0, lambda = -1 so (lambda(y))^2+0=289
which is -y^2=289, y = -17

So why do both the lexicographically first points contain negative coordinates?

I just rambled a whole lot. Basically if someone could give me an explanation for why the correct answer is correct that'd be great.

Thanks a lot,
Gregg
 
Physics news on Phys.org
  • #2
glid02 said:
Hey, I need help with a problem involving Lagrange multipliers...

Here is the question:
Find the absolute maximum and minimum of the function
f(x,y) = x^2-y^2 subject to the constraint x^2+y^2=289.

As usual, ignore unneeded answer blanks, and list points in lexicographic order.

I found that the max is 289 and the min is -289.

I took the gradients of each and came up with:
2x-2y = lambda2x+lambda2y
[tex]2x \vec{i}- 2y\vec{j}= \lambda 2x\vec{i}+ \lambda 2y\vec{j}[/tex]
Dont forget that these are vector so that your next equations are true:

x=lambda(x)
-y=lambda(y)

Here's where I'm stuck, I know lambda = +/-1, but I can't come up with an equation to solve for x or y in terms of y or x. The best idea I can come up with is to solve for lambda from -y=lambda(y), which is just
-y/y and plug it into x=lambda(x), but that doesn't help any when solving for y.

I guess lambda = 1 when y = 0 and lambda = -1 when x = 0.
Okay, good. x= lambda x gives immediately that either x= 0 or lamba= 1. If x= 0, then, since x^2+y^2=289, y^2= 289, y= 17 or -17.
If lambda= 1, y must be 0 so x^2= 289 and x= 17 or -17.

y= -lambda y gives immediately that either

OK, I found the correct answers, but I could still use an explanation as to how to list them lexicographically.

The correct answer is:
max = 289 at points (-17, 0) and (17,0)
min = -289 at points (0,-17) and (0,17)

When y = 0, lambda = 1 so (lambda(x))^2+0=289
which is x^2=289, x = 17

When x = 0, lambda = -1 so (lambda(y))^2+0=289
which is -y^2=289, y = -17

So why do both the lexicographically first points contain negative coordinates?

I just rambled a whole lot. Basically if someone could give me an explanation for why the correct answer is correct that'd be great.

Thanks a lot,
Gregg
I'm not at all sure why you are worried about listing them "lexicographically". If the problem just asks you to find maximum and minimum, then it doesn't matter how you order them.

To answer your question about "why do both the lexicographically first points contain negative coordinates?" (which has nothing to do with "maximization" or "Lagrange"), remember that "lexicographic" ordering is derived from alphabetical ordering of words: look at the first letter and order by that. If both words have the same first letter, look at the second letter and so on. In "lexicographic" ordering of ordered pairs of numbers, look at the first number and order numerically by that. Of course, pairs having first number negative will come before pairs having first number positive. If both first numbers are the same (both 0 for example), then order by the second member. Again, negative numbers are "less than" positive numbers and will come first.
 

What is the Lagrange multipliers problem?

The Lagrange multipliers problem is a mathematical optimization problem that involves finding the maximum or minimum value of a function subject to one or more constraints. It is named after the mathematician Joseph-Louis Lagrange.

What is the use of Lagrange multipliers?

Lagrange multipliers are used to solve constrained optimization problems in which the objective function and constraints are differentiable. They help in finding the optimal values of the variables that satisfy all the constraints.

What is the intuition behind Lagrange multipliers?

The intuition behind Lagrange multipliers is that at the optimal solution, the gradient of the objective function and the gradients of the constraints are parallel. This means that the optimal solution is found at the point where these gradients are equal.

How do you solve a Lagrange multipliers problem?

To solve a Lagrange multipliers problem, you first need to set up the Lagrangian function, which is the objective function plus the product of the Lagrange multipliers and the constraints. Then, you take the partial derivatives of the Lagrangian function with respect to each variable and set them equal to zero. This will give you a system of equations that can be solved to find the optimal values of the variables and the Lagrange multipliers.

What are some real-life applications of Lagrange multipliers?

Lagrange multipliers have applications in various fields such as economics, engineering, and physics. They are used to optimize production processes, maximize profits in businesses, and solve problems in mechanics and thermodynamics. They are also used in machine learning and data analysis to find optimal solutions to complex problems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
454
  • Calculus and Beyond Homework Help
Replies
10
Views
301
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
822
  • Calculus and Beyond Homework Help
Replies
6
Views
838
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
270
Back
Top