Lagrange method to find extremes

In summary, the equation x² + 2y² ≤ 1 defines an inequality between the two sides of the equation. When the inequality is strict, meaning that there is no boundary condition on x or y, the two extrema are (0,0) and (-1,0). However, when the inequality is exact, meaning that the boundary condition on x and y is (x,y) = (-1,0), then the two extrema are (-6/√38,-6/√38).
  • #1
LCSphysicist
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Homework Statement
Study the function's maximum and minimum
Relevant Equations
All below.
1597146354939.png

ƒ(x,y) = 3x + y
x² + 2y² ≤ 1

It is easy to find the maximum, the really problem is find the minimum, here is the system:

(3,1) = λ(2x,4y)
x² + 2y² ≤ 1

how to deal with the inequality?
 
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  • #2
Sorry, I find the picture a little unclear. f(x,y)=3x+y is the function you're maximizing/minimizing, and ##x^2+2y^2\leq1## is the region of the plane you're doing it in?

The extrema all are in one of two states:
1.) The inequality is strict. In that case there's no boundary condition, and the gradient needs to just be zero.

2.) The inequality is exact. In that case you can use Lagrange multipliers to figure out where the extrema are.

I think you need to just solve these two cases separately.
 
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  • #3
Office_Shredder said:
Sorry, I find the picture a little unclear. f(x,y)=3x+y is the function you're maximizing/minimizing, and ##x^2+2y^2\leq1## is the region of the plane you're doing it in?

The extrema all are in one of two states:
1.) The inequality is strict. In that case there's no boundary condition, and the gradient needs to just be zero.

2.) The inequality is exact. In that case you can use Lagrange multipliers to figure out where the extrema are.

I think you need to just solve these two cases separately.
The gradients i found can never be zero, that is:

∇ƒ = (3,1,0)
so what to do?

I though wrongly that it would be, for example, (0,0) the minimum, but we can find to (-1,0) and so f = -3.
 
  • #4
If the gradient is never zero, then all the extrema must be where the inequality is an equality, and you can use Lagrange multipliers.
 
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  • #5
Office_Shredder said:
If the gradient is never zero, then all the extrema must be where the inequality is an equality, and you can use Lagrange multipliers.
Okay, that's good to know, i found:

1597336861797.png


the book:

1597336589759.png

but (-6/√38,-6/√38) don't satisfy!
(-6/√38)² + 2*(-6/√38)² ≤ 1

Maybe am i right?
 

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  • #6
I think you got it right, and the book has a typo.
 
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1. What is the Lagrange method to find extremes?

The Lagrange method is a mathematical technique used to find the maximum or minimum value of a function subject to certain constraints. It involves setting up a system of equations and solving for the critical points of the function.

2. How does the Lagrange method work?

The Lagrange method uses the concept of constrained optimization, where the maximum or minimum value of a function is found within a specific set of constraints. It involves setting up a system of equations using the original function and the constraints, and then solving for the critical points using the method of Lagrange multipliers.

3. When is the Lagrange method used?

The Lagrange method is commonly used in optimization problems in various fields such as economics, engineering, and physics. It is particularly useful when the constraints are difficult to incorporate into the original function or when the function is multivariate.

4. What are the advantages of using the Lagrange method?

One of the main advantages of the Lagrange method is that it allows for the optimization of a function subject to multiple constraints. It also provides a systematic approach to solving constrained optimization problems and can be applied to both single-variable and multi-variable functions.

5. Are there any limitations to the Lagrange method?

One limitation of the Lagrange method is that it may not always provide the global maximum or minimum of a function, but rather local extrema. Additionally, it can become computationally intensive for functions with a large number of variables and constraints.

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