Max/Min of f Using Lagrange Multipliers

In summary, To find the maximum and minimum of f=-x^2+y^2 about the ellipse x^2+4y^2=4, the values of x and y can be determined by plugging in lambda=-1 and lambda=1/4 into the given equations and solving for the corresponding values.
  • #1
alejandrito29
150
0
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]?
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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 :(
 
  • #4
What is it you don't understand??

What does the second equation look like if you insert lambda=-1?
 
  • #5
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 ?
 
  • #6
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.
 
  • #7
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
 

Related to Max/Min of f Using Lagrange Multipliers

What is the concept of Max/Min of f Using Lagrange Multipliers?

The concept of Max/Min of f Using Lagrange Multipliers is a method for finding the maximum or minimum value of a function with given constraints. It involves using the Lagrange multiplier, which is a mathematical tool used to optimize multivariable functions subject to constraints.

When is Max/Min of f Using Lagrange Multipliers used?

Max/Min of f Using Lagrange Multipliers is used when there are constraints present in a multivariable function and the objective is to find the maximum or minimum value of the function. It is commonly used in optimization problems in mathematics and physics.

What are the steps involved in finding the Max/Min of f Using Lagrange Multipliers?

The steps involved in finding the Max/Min of f Using Lagrange Multipliers are:

  1. Identify the function to be optimized and the constraints.
  2. Formulate the Lagrangian function by multiplying the constraints with a Lagrange multiplier.
  3. Take the partial derivatives of the Lagrangian function with respect to each variable.
  4. Set the partial derivatives equal to 0 and solve the resulting system of equations.
  5. Verify if the solution satisfies the constraints and find the corresponding value of the function.

What are the advantages of using Max/Min of f Using Lagrange Multipliers?

Some advantages of using Max/Min of f Using Lagrange Multipliers are:

  • It can be used to find the maximum or minimum value of a function with constraints, which cannot be found using traditional optimization methods.
  • It provides a systematic approach to solving optimization problems with constraints.
  • It can be applied to a wide range of functions and constraints.

What are the limitations of Max/Min of f Using Lagrange Multipliers?

Some limitations of Max/Min of f Using Lagrange Multipliers are:

  • It can only find local maxima or minima, not global ones.
  • The Lagrangian function can become complicated for functions with multiple constraints.
  • The method may not always provide a solution, or the solution may be difficult to interpret.

Similar threads

I Can Lagrange multipliers be used to find a function?
    • Calculus
Replies
13
Views
1K
I The Euler-Lagrange equation and the Beltrami identity
    • Calculus
Replies
1
Views
1K
MHB 2.6.5 Implicit differentiation
    • Calculus
Replies
1
Views
1K
MHB Optimization - Lagrange multipliers : minimum cost/maximum production
    • Calculus
Replies
3
Views
1K
Unsolvable max/min of surface with constraint
    • Calculus and Beyond Homework Help
Replies
9
Views
1K
How should I find the nontrivial stationary paths?
    • Calculus and Beyond Homework Help
Replies
10
Views
532
I Lagrange Multiplier where constraint is a rectangle
    • Calculus
Replies
4
Views
3K
I Result of this integral in large Lambda limit
    • Calculus
Replies
12
Views
1K
MHB Solve Lagrange Multipliers Problem with x-4y=1 Constraint
    • Calculus
Replies
1
Views
1K
How to derive the associated Euler-Lagrange equation?
    • Calculus and Beyond Homework Help
Replies
6
Views
899

Hot Threads

Recent Insights

Back
Top