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

[GWR309]

f(x,y)=x^2y with the constraint of x^2+2y^2=6

Use lagrange multipliers to find the extrema.

Thanks!

 

[Prove It]

Well, have you tried to solve the system

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

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

 

[GWR309]

Well, have you tried to solve the system

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

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

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

 

[MarkFL]

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)

 

[Prove It]

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 [math]\displaystyle \nabla f(x, y) = \left( \frac{\partial f }{\partial x} , \frac{\partial f}{\partial y} \right)[/math]

 

[MarkFL]

The way I learned to use Lagrange multipliers, while Prove It is being more rigorous, is to write:

The objective function is:

$$f(x,y)=x^2y$$

subject to the constraint:

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

Now, first find the implications of the system:

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

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

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

 

[Prove It]

The way I learned to use Lagrange multipliers, while Prove It is being more rigorous, is to write:

The objective function is:

$$f(x,y)=x^2y$$

subject to the constraint:

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

Now, first find the implications of the system:

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

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

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

[math]\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*}[/math]