Consider the ellipse with equation x2 + 2xy +4y2 = 9.

[thumbsupfor]

Hi guys, I'm really stuck with this question and I'm not sure where to start. I know that I need two parts/ functions to be able to solve this but haven't seen anywhere online on how to do so. Help would be much appreciated, cheers.

Consider the ellipse with equation x2 + 2xy +4y2 = 9.
Using the method of Lagrange multipliers, find the lowest point on the ellipse (that is, the point with the smallest y-coordinate).

 

[khansaheb]

Hi guys, I'm really stuck with this question and I'm not sure where to start. I know that I need two parts/ functions to be able to solve this but haven't seen anywhere online on how to do so. Help would be much appreciated, cheers.

Consider the ellipse with equation x2 + 2xy +4y2 = 9.
Using the method of Lagrange multipliers, find the lowest point on the ellipse (that is, the point with the smallest y-coordinate).

With these types of questions, I plot the function (using any graphical software ), to have an "idea" about the answer.

I don't know what do you mean by

I need two parts/ functions to be able to solve this

My interpretation of the problem is:

Calculate the local minimum of f(x,y) = x^2 + 2*x*y + 4*y^2 - 9 , using Lagrange multiplier.

Have you solved any example problem in that topic?

 

[Dr.Peterson]

Hi guys, I'm really stuck with this question and I'm not sure where to start. I know that I need two parts/ functions to be able to solve this but haven't seen anywhere online on how to do so. Help would be much appreciated, cheers.

Consider the ellipse with equation x2 + 2xy +4y2 = 9.
Using the method of Lagrange multipliers, find the lowest point on the ellipse (that is, the point with the smallest y-coordinate).

I think the "two functions" you refer to are the f and g described here:

Theorem 14.8.1: Let f and g be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve g(x,y)=0. Suppose that f, when restricted to points on the curve g(x,y)=0, has a local extremum at the point (x₀,y₀) and that ∇g(x₀,y₀)≠0. Then there is a number λ called a Lagrange multiplier, for which ∇f(x₀,y₀)=λ∇g(x₀,y₀).​

You have to minimize the function f(x,y) = y subject to the constraint g(x,y) = x² + 2xy + 4y² - 9 = 0.

Does that answer your question?

 

[lookagain]

Consider the ellipse with equation x2 + 2xy +4y2 = 9.

You have x^2 + 2xy + 4y^2 = 9 \(\displaystyle \ \ \) or \(\displaystyle \ \ x^2 + 2xy + 4y^2 = 9.\)

You don't need to use Lagrange multipliers to solve this. You can set it up as a quadratic equation with y as the variable:

\(\displaystyle 4y^2 +(2x)y + (x^2 - 9) = 0 \ \ \) Then, you can use the Quadratic Formula on it where A = 4, B = 2x, and C = (x^2 - 9).

Take the derivative of that expression, which is in terms of x, set it equal to 0, and solve for an appropriate x-value.

Then, substitute that x-value into the y-value expression which gives the lower value.

 

[Bruce]

solve for an appropriate x-value

I tried something similar to your way and also took derivative but there was only one root so not sure what numbers might be inappropriate.

You don't need to use Lagrange multiplier

Yeah, I can't remember Lagrange but I forgot a lot from college (stuff never used again). Your way seems basic, but don't students "need" to do what teacher says?

 

[khansaheb]

I tried something similar to your way and also took derivative but there was only one root so not sure what numbers might be inappropriate.
Yeah, I can't remember Lagrange but I forgot a lot from college (stuff never used again). Your way seems basic, but don't students "need" to do what teacher says?

Follow the advice given to you in response #2 and #3.

 

[Dr.Peterson]

Follow the advice given to you in response #2 and #3.

Except it wasn't given to him. Bruce is not the OP.

I tried something similar to your way and also took derivative but there was only one root so not sure what numbers might be inappropriate.


Yeah, I can't remember Lagrange but I forgot a lot from college (stuff never used again). Your way seems basic, but don't students "need" to do what teacher says?

Responding to a thread that isn't yours in this way isn't helpful, and can be confusing. Please don't do that.

 

[Dr.Peterson]

I tried something similar to your way and also took derivative but there was only one root so not sure what numbers might be inappropriate.

If you'd like help with your attempt, you can show us your work, and maybe we can point out what you are missing. I can't tell what you mean by "only one root". (But I do think that method, if I understand it correctly, is considerably harder than other ways.)

Your way seems basic, but don't students "need" to do what teacher says?

Yes, assuming the problem was quoted exactly, Lagrange is the required method. And if the OP ever responds (or if you were to look up the method and show an attempt of your own), then we might be able to discuss the details. It is pretty straightforward using that method.

But it can also be interesting to think about alternative methods, whether they are easier or harder than what is being taught. A good student will stop after any interesting problem and ask both "what have I learned by doing this?" and also "could I have done this a different way?" By trying other ways, you gain a fuller understanding of what methods work under what circumstances.

My own alternative method was just to differentiate implicitly and set that to 0 to find the minimum value of y. That's very easy.

It could also be solved without calculus, by using a discriminant to find what horizontal line intersects the curve only once.

 

[Bruce]

I can't tell what you mean by "only one root"

oh I just commented because the other suggestion to set derivative=0 and "solve for appropriate x-value" was curious because I see only 1 possible x-value. No worries (maybe they worked something different than I did).

I accept earlier critisism about not posting in another's thread. Sorry for the confusion.

And I will think about your discriminant method, thanks for new suggestion.

 

[Dr.Peterson]

oh I just commented because the other suggestion to set derivative=0 and "solve for appropriate x-value" was curious because I see only 1 possible x-value. No worries (maybe they worked something different than I did).

I tried something similar to your way and also took derivative but there was only one root so not sure what numbers might be inappropriate.

What he said was,

Take the derivative of that expression, which is in terms of x, set it equal to 0, and solve for an appropriate x-value.

I suspect you are misunderstanding the word "appropriate". It doesn't necessarily mean one of several possible values (of a root, for example), though it may. I take it to mean "solve for the value of x that accomplishes the goal" -- which is the one that makes y a minimum (rather than, say, a maximum).

I'm not sure of the details of the intended method, but the quadratic formula includes a "[imath]\pm[/imath]", and therefore so does the derivative; so in fact there are two possibilities, possibly corresponding to the max and the min. (In fact, that will go away, and there will be two values for a different reason!) Again, if you showed your work (which you only said was similar to what he described), there is probably more to be said.