Optimization using Lagrange multipliers

In summary, the problem involves finding the maximum and minimum values of two functions, f(x,y) = x^2 + y^2 and g(x,y) = x^4 + y^4, using the Lagrange multiplier method. After calculating the gradient of both functions and setting them equal to each other, it is found that x^2 = y^2 and x^4 + y^4 = 2. This leads to four possible points: (1, 1), (-1, -1), (1, -1), and (-1, 1). The maximum value is 2, achieved at (1, 1) and (-1, -1). To find the minimum value, the correct
  • #1
tinkus
13
0
1. Homework Statement [/b]

f[tex]\left(x,y\right)[/tex] = x^2 +y^2
g[tex]\left(x,y\right)[/tex] = x^4+y^4 = 2
Find the maximum and minimum using Lagrange multiplier

Homework Equations





The Attempt at a Solution



grad f = 2xi +2yj
grad g= 4x^3i + 4y^3j

grad f= λ grad g
2x=4x^3λ and 2y= 4y^3λ
2x^2 = 2y^2
x^2=y^2
x= [tex]\pm[/tex]y
x^4+x^4=2
x=y= [tex]\pm1[/tex]
max= 1+1=2 @ [tex]\left(1,1\right)[/tex] and [tex]\left(-1,-1\right)[/tex]

I don't know how to find the min and not sure about the max above



 
Last edited:
Physics news on Phys.org
  • #2


tinkus said:
1. Homework Statement [/b]

f[tex]\left(x,y\right)[/tex] = x^2 +y^2
g[tex]\left(x,y\right)[/tex] = x^4+y^4
Find the maximum and minimum using Lagrange multiplier
My recomendation is that you go back and read the problem carefully! What you have written here makes no sense. Usually you use Lagrange multiplier method maximize or minimize a function subject to some constraint. You have two functions with no constraint. Is one of those, either f or g, supposed to be equal to a number?

Homework Equations





The Attempt at a Solution



grad f = 2xi +2yj
grad g= 4x^3i + 4y^3j

grad f= λ grad g
2x=4x^3λ and 2y= 4y^3λ
2x^2 = 2y^2
x^2=y^2
x= [tex]\pm[/tex]y
x^4+x^4=2
x=y= [tex]\pm1[/tex]
max= 1+1=2 @ [tex]\left(1,1\right)[/tex] and [tex]\left(-1,-1\right)[/tex]

I don't know how to find the min and not sure about the max above
 
  • #3


yes the constraint function is incorrect, i ommitted =2
 
  • #4


You have found that y2= x2 and you know that [itex]x^4+ y^4= 2[/itex]. That tells you that [itex]x= \pm 1[/itex] and [itex]x= \pm 1[/itex]. That gives you four possible points: (1, 1), (-1, -1), (1, -1), and (-1, 1). You might want to consider whether there are both maximum and mimimum values.
 
  • #5


the points all equal 2(max). i still need to find the min which according to the answer key is sqrt2, I don't know how to get that.
 

1. What is optimization using Lagrange multipliers?

Optimization using Lagrange multipliers is a method used to find the maximum or minimum value of a function subject to certain constraints. It involves using Lagrange multipliers, which are mathematical tools that help incorporate the constraints into the optimization problem.

2. When is optimization using Lagrange multipliers used?

Optimization using Lagrange multipliers is commonly used in engineering, physics, economics, and other fields where there is a need to optimize a function while satisfying certain constraints. It is also used in multivariable calculus and optimization courses.

3. How does optimization using Lagrange multipliers work?

The method involves setting up a Lagrangian function, which is the original function being optimized along with the constraints multiplied by Lagrange multipliers. The Lagrangian function is then differentiated with respect to the variables and the Lagrange multipliers to find the critical points, which are the potential solutions. These critical points are then evaluated to find the optimal solution.

4. What are the advantages of using optimization using Lagrange multipliers?

One advantage is that it allows for optimization with constraints to be solved using techniques from calculus, which is more familiar to many scientists and engineers. Additionally, it can handle both equality and inequality constraints. It also provides a way to find the optimal solution without having to consider every possible combination of constraints, which can be time-consuming.

5. Are there any limitations to optimization using Lagrange multipliers?

One limitation is that it can only be used for continuous functions and constraints. It also may not always provide the global optimum, as it only finds the critical points of the Lagrangian function. Additionally, it can be challenging to use with a large number of constraints or variables.

Similar threads

  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
467
  • Calculus and Beyond Homework Help
Replies
6
Views
849
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
834
  • Calculus and Beyond Homework Help
Replies
2
Views
540
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
874
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
504
Back
Top