Find all critical and stationary points: Legrange multipliers

[mikeyrichster]

Find all critical and stationary points of the function f(x,y)=x^3-y^2 subject to the inequality constraint c(x,y)=1-x^2-y^2 >=0

So far I've deduced that I need to use a lagrange multiplier L say, so i think i need to solve the equations :
3x^2=-2Lx
-2y^2=-2Ly

and

1-x^2-y^2 >=0

Is that correct? I'm finding it hard to solve as the 1st and second EQ only have one variable other than the multiplier..

any help would be appreciated!

 

[Je m'appelle]

so you want to find the critical points of

$$f(x,y) = x^3 - y^2$$

Subject to the following constraint

$$C(x,y) = 1 - x^2 - y^2 \geq 0$$

Through Lagrange Multipliers?

First of all, you should remember from the definition of Lagrange Multipliers that

$$\bigtriangledown f(x,y) = \lambda \bigtriangledown C(x,y)$$

And

$$C(x,y) = k$$

Notice in this case that we have from the constraint

$$1 - x^2 - y^2 \geq 0$$

$$x^2 + y^2 \leq 1$$

And what does that equation resembles? That's right, a circle (not a circumference, a circumference would have been a curve, which would have happened if there were an equality sign instead of an inequality sign, as a circle is a group of several points limited by the central point and the circumference) of radius 1.

Then you'll have an easily solvable system

$$\frac{\partial f}{\partial x} = 3x^2 = \lambda\frac{\partial C}{\partial x} = 2x\lambda$$

$$\frac{\partial f}{\partial y} = -2y = \lambda\frac{\partial C}{\partial y} = 2y\lambda$$

$$0 \leq x^2 + y^2 \leq 1$$

 

[mikeyrichster]

Hey thanks for your help that's brilliant!

So I have the solvable system of EQ's:

-2y = 2y$$\lambda$$

and

$$3x^2$$=2x$$\lambda$$

Am I correct in saying now that x = -2/3 and y = +/-sqrt 13/9 ? sorry about my lack of LaTeX typesetting.

:)

 

[Je m'appelle]

Hello there little man, I'm sorry for the delay but I've been busy, I don't know if you still need this or if you have already solved, but anyway, I'll leave it here just in case.

Hey thanks for your help that's brilliant!

So I have the solvable system of EQ's:

-2y = 2y$$\lambda$$

and

$$3x^2$$=2x$$\lambda$$

Yes, this is perfect.

Am I correct in saying now that x = -2/3 and y = +/-sqrt 13/9 ?

Hum, let's check it together, shall we?

We've seen that

$$-2y = 2y\lambda ,\ so\ \lambda = -1$$

Look that $$y$$ is not defined yet, we just found $$\lambda$$, so we'll find $$y$$ by first finding $$x$$.

$$3x^2 = 2x\lambda,\ where\ \lambda = -1$$

$$3x^2 = -2x$$

$$3x^2 + 2x = 0$$

$$x(3x + 2) = 0$$

So we'll have two solutions for $$x$$

$$x = 0\ and\ 3x = -2,\ x = \frac{-2}{3}$$

Now we'll have to plug both values found for $$x$$ on the constraint equation in order to find $$y$$

From the constraint (notice that we won't need anymore the inequality sign by now) we have

$$x^2 + y^2 = 1$$

For $$x = 0$$,

$$(0)^2 + y^2 = 1,\ therefore\ y = \pm 1$$

For,

$$x = \frac{-2}{3}$$

$$(\frac{-2}{3})^2 + y^2 = 1$$

$$\frac{4}{9} + y^2 = 1$$

$$4 + 9y^2 = 9$$

$$9y^2 = 5,\ then\ y = \pm \frac{\sqrt{5}}{3}$$


Then the critical points are

$$P_1 = (0,1)\ and\ P_2 = (0, -1)\ and\ P_3 = (\frac{-2}{3},\frac{\sqrt{5}}{3})\ and\ P_4 = (\frac{-2}{3},\frac{-\sqrt{5}}{3})\ and\ P_5 = (0,0)$$

Remember that $$P_5 = (0,0)$$ is a critical point inside the disk once it satisfies the definition for a critical point of $$f(x,y)$$ that is

$$\frac{\partial f}{\partial x} = 0$$

$$\frac{\partial f}{\partial y} = 0$$

Therefore,

$$\frac{\partial f}{\partial x} = 3x^2 = 0,\ then\ x = 0$$

$$\frac{\partial f}{\partial y} = -2y = 0,\ then\ y = 0$$

So, $$P_5 = (0,0)$$ is a critical point by definition.

sorry about my lack of LaTeX typesetting.

No problem at all, I understood it perfectly.

Just in case you're curious about learning $$LaTeX$$, there's this introductory topic on PF:

https://www.physicsforums.com/showthread.php?t=8997

It's pretty easy to learn, you can master the basics in matter of minutes, so check it out.

See you little man, and good luck with those studies.

 

[Dickfore]

Your contstraint is not binding. Therefore, straighforward application of Lagrange's multipliers is not applicable.

First, you need to find if $$f(x, y)$$ has any stationary points inside the region of interest. Then, you should also find if the function has critical points on the boundary of the region you are considering. Only in this step do you use Lagrange multipliers (because the boundary is defined by the equality sign in the constraint). Nevertheless, you might notice that the unconstrained problem is equal to the constrained with the value $\lambda = 0$ for the Lagrange's multiplier, just to avoid extra algebra.

 

[mikeyrichster]

Thanks for all your help, I've finally got my head around this... and cheers for the LaTeX heads up. I've spent ages looking for good LaTeX resources for beginners.

:)