Max/Min of f Using Lagrange Multipliers

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Discussion Overview

The discussion revolves around finding the maximum and minimum values of the function f = -x² + y² constrained by the ellipse defined by the equation x² + 4y² = 4. Participants are exploring the application of Lagrange multipliers to solve this problem.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant presents the problem and their initial equations derived from applying Lagrange multipliers, leading to two possible values for lambda: -1 and 1/4.
  • Another participant prompts further exploration by asking what implications arise for y when lambda = -1 is substituted into the second equation.
  • There is a request for clarification on the implications of the equations, indicating some confusion about the steps involved in solving for x and y.
  • Further responses guide the original poster to substitute lambda values back into the equations to find corresponding x and y values, but the understanding of these steps remains unclear for some participants.
  • One participant reiterates the process of substituting lambda values into the equations to find specific values for x and y, confirming the earlier steps taken.

Areas of Agreement / Disagreement

Participants appear to be working through the problem collaboratively, but there is no consensus on the clarity of the steps involved, as some express confusion about the implications of their calculations.

Contextual Notes

Some participants may be missing assumptions about the behavior of the equations or the implications of the values of lambda. The discussion does not resolve the mathematical steps or clarify all uncertainties expressed.

alejandrito29
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In a exercise says:

Find max a min of [tex]f=-x^2+y^2[/tex] abaut the ellipse [tex]x^2+4y^2=4[/tex]

i tried [tex]-2x=\lambda 2x[/tex]
[tex]2y=\lambda 8y[/tex]
[tex]x^2+4y^2-4=0[/tex]

then [tex]\lambda =-1[/tex] or [tex]\lambda =\frac{1}{4}[/tex] , but, ¿how i find [tex]x,y[/tex]?
 
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Well, suppose you identically fulfill your first eqaution by choosing lambda=-1.

What does this imply that "y" must equal, by looking at your second equation?
 
arildno said:
Well, suppose you identically fulfill your first eqaution by choosing lambda=-1.

What does this imply that "y" must equal, by looking at your second equation?

thank, but, i don't understand :(
 
What is it you don't understand??

What does the second equation look like if you insert lambda=-1?
 
Okay you seem to be going along fine. From your first equation you determined that [tex]\lambda = -1[/tex] So now if you plug [tex]\lambda = -1[/tex] into your second equation what must y be ? and when you plug that into your third equation what do you get for x ? Now from your second equation you determined [tex]\lambda = \frac{1}{4}[/tex] so when you plug that into your first equation what must x be ? and then what do you get for y when you plug into your third equation ?
 
alejandrito29 said:
In a exercise says:

Find max a min of [tex]f=-x^2+y^2[/tex] abaut the ellipse [tex]x^2+4y^2=4[/tex]

i tried [tex]-2x=\lambda 2x[/tex]
[tex]2y=\lambda 8y[/tex]
[tex]x^2+4y^2-4=0[/tex]

then [tex]\lambda =-1[/tex] or [tex]\lambda =\frac{1}{4}[/tex] , but, ¿how i find [tex]x,y[/tex]?
If [itex]\lambda= -1[/itex] then the second equation becomes [itex]2y= -8y[/itex] so that [itex]y= 0[/itex]. You can solve [tex]x^2= 4[/tex] for the corresponding x values.

If [itex]\lambda= \frac{1}{4}[/itex], then the first equation becomes [itex]-2x= (1/2)x[/itex] so that [itex]x= 0[/itex]. Solve [itex]4y^2= 4[/itex] for y.
 
HallsofIvy said:
If [itex]\lambda= -1[/itex] then the second equation becomes [itex]2y= -8y[/itex] so that [itex]y= 0[/itex]. You can solve [tex]x^2= 4[/tex] for the corresponding x values.

If [itex]\lambda= \frac{1}{4}[/itex], then the first equation becomes [itex]-2x= (1/2)x[/itex] so that [itex]x= 0[/itex]. Solve [itex]4y^2= 4[/itex] for y.

Very Thanks
 

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