Lagrange Multipliers: Find Extrema of f(x,y)=x^2y

In summary, the gradient vectors for the two functions are not the same, and you will need to find the critical points to find the extrema.
  • #1
GWR309
4
0
f(x,y)=x^2y with the constraint of x^2+2y^2=6

Use lagrange multipliers to find the extrema.

Thanks!
 
Physics news on Phys.org
  • #2
Well, have you tried to solve the system

\(\displaystyle \displaystyle \begin{align*} \nabla f(x, y) &= \lambda \nabla g(x, y) \\ g(x, y) &= k \end{align*}\)

yet? Here \(\displaystyle \displaystyle f(x, y) = x^2y\) and \(\displaystyle \displaystyle g(x, y) = x^2 + 2y^2 = 6\).
 
  • #3
Prove It said:
Well, have you tried to solve the system

\(\displaystyle \displaystyle \begin{align*} \nabla f(x, y) &= \lambda \nabla g(x, y) \\ g(x, y) &= k \end{align*}\)

yet? Here \(\displaystyle \displaystyle f(x, y) = x^2y\) and \(\displaystyle \displaystyle g(x, y) = x^2 + 2y^2 = 6\).

Yeah I tried. I ended up with 2y^2+sqrt(2)y-6 which doesn't seem right and if it is right, I don't know how to solve it
 
  • #4
GWR309 said:
f(x,y)=x^2y with the constraint of x^2+2y^2=6

Use lagrange multipliers to find the extrema.

Thanks!

Hello again, GWR309!(Wave)

When I and others bring questions from other sites here, we provide a full solution, as a means of increasing our knowledge base and to demonstrate to guests the type of expertise available here at MHB.

As a registered member, you will now be encouraged to show what you have tried so that we may help you be a part of the learning process by taking part in getting to the solution. It would actually be lazy of us to provide full solutions to everyone, and would not meet our goal of teaching rather than simply providing answers, which is of minimal benefit to students.

We will provide suggestions/hints, and then expect you to either give feedback on your progress, or to ask for further clarification. We will continue until you have solved the problem, and you will have learned much more and will gain a sense of accomplishment in having actually taken part in finding the solution.

I just wanted to post that bit of information so that you understand why I worked out your problem from Yahoo! Answers in full when we normally try to actively engage students on the solutions process. I will now leave you in Prove It's very capable hands. (Cool)
 
  • #5
GWR309 said:
Yeah I tried. I ended up with 2y^2+sqrt(2)y-6 which doesn't seem right and if it is right, I don't know how to solve it

Sorry but that's not even close. Start by evaluating the gradient vector for each of those functions. Remember that \(\displaystyle \displaystyle \nabla f(x, y) = \left( \frac{\partial f }{\partial x} , \frac{\partial f}{\partial y} \right)\)
 
  • #6
The way I learned to use Lagrange multipliers, while Prove It is being more rigorous, is to write:

The objective function is:

\(\displaystyle f(x,y)=x^2y\)

subject to the constraint:

\(\displaystyle g(x,y)=x^2+2y^2-6=0\)

Now, first find the implications of the system:

\(\displaystyle f_x(x,y)=\lambda g_x(x,y)\)

\(\displaystyle f_y(x,y)=\lambda g_y(x,y)\)

Then use the implications in the constraint to find the critical points. Can you write down the system from which to take the implications?
 
  • #7
MarkFL said:
The way I learned to use Lagrange multipliers, while Prove It is being more rigorous, is to write:

The objective function is:

\(\displaystyle f(x,y)=x^2y\)

subject to the constraint:

\(\displaystyle g(x,y)=x^2+2y^2-6=0\)

Now, first find the implications of the system:

\(\displaystyle f_x(x,y)=\lambda g_x(x,y)\)

\(\displaystyle f_y(x,y)=\lambda g_y(x,y)\)

Then use the implications in the constraint to find the critical points. Can you write down the system from which to take the implications?

Which if we wrote as vectors would look like

\(\displaystyle \displaystyle \begin{align*} \left[ \begin{matrix} f_x (x, y) \\ f_y (x, y) \end{matrix} \right] &= \lambda \left[ \begin{matrix} g_x(x, y) \\ g_y (x, y) \end{matrix} \right] \\ \nabla f(x, y) &= \lambda \nabla g(x, y) \end{align*}\)
 

Related to Lagrange Multipliers: Find Extrema of f(x,y)=x^2y

1. What are Lagrange multipliers?

Lagrange multipliers are a mathematical tool used to find the extrema (maximum or minimum) of a function subject to one or more constraints.

2. How do Lagrange multipliers work?

To use Lagrange multipliers, we first set up an equation involving the function we want to optimize (in this case, f(x,y)=x^2y) and the constraints. We then use the method of partial derivatives to solve for the values of x and y that will give us the extrema.

3. Why do we use Lagrange multipliers?

Lagrange multipliers are useful because they allow us to find the extrema of a function subject to constraints without having to solve a system of equations.

4. Can Lagrange multipliers be used with more than two variables?

Yes, Lagrange multipliers can be used with any number of variables. However, as the number of variables increases, the equations become more complex and may be more difficult to solve.

5. What are some real-world applications of Lagrange multipliers?

Lagrange multipliers are commonly used in economics, physics, and engineering to optimize functions subject to constraints. For example, they can be used to determine the most efficient way to allocate resources or to design structures that can withstand certain forces.

Similar threads

Replies
13
Views
1K
Replies
1
Views
1K
Replies
3
Views
1K
  • Calculus
Replies
9
Views
2K
Replies
4
Views
3K
Replies
9
Views
2K
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
513
Replies
2
Views
951
Back
Top