Lagrangian Multipliers to find maximum and minimum values

In summary, the Lagrange multiplier method states that the maximum or minimum value of a function will occur when the gradients of the function and the constraint are scalar multiples of each other. However, in cases where the constraint curve is a straight line and passes through the center of the level curve of the function, the gradients are never parallel. This means that the constraint does not actually constrain the function and the maximum or minimum value will be where the gradient of the function is 0. Similarly, if the constraint curve intersects the level curve of the function at a tangent, the supposed maximum or minimum may occur at a random point.
  • #1
cooev769
114
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I'm just learning this theory and the maths is really trivial but the theory is slightly confusing me.

I understand that if we have some function z=f(x,y) and we graph this on a three dimensional set of axis we will have some surface, we can then extend this by creating level curves in the x, y plane for differing values of z.

We then set up a function g(x,y)=k such that k is a constant, we call this our constraint curve and it looks something as shown in the link I have provided.

http://en.wikipedia.org/wiki/File:LagrangeMultipliers2D.svg

I understand how the theory that the maximum/minimum value will occur when the grad of function f and the grad of function g are scalar multiples of each other. But imagine for a second that the constraint curve was a straight line and it went straight through the centre of the circle created by the level curve of the function f. In that case the maximum value would be at the centre of this level curve, but to my knowledge at no time were the gradients parallel.

Or imagine a sloping flat surface sloping up. If the constraint curve started off running up the slope then for a second ran tangent to one of the level curves and then ran at some angle up this surface and stopped, then the point at which the gradients were scalar multiples only in the middle of the curves, so the supposed maximum/minimum occurred at a random place. If you could clarify this it would be extremely helpful thanks!
 
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  • #2
cooev769 said:
I'm just learning this theory and the maths is really trivial but the theory is slightly confusing me.

I understand that if we have some function z=f(x,y) and we graph this on a three dimensional set of axis we will have some surface, we can then extend this by creating level curves in the x, y plane for differing values of z.

We then set up a function g(x,y)=k such that k is a constant, we call this our constraint curve and it looks something as shown in the link I have provided.

http://en.wikipedia.org/wiki/File:LagrangeMultipliers2D.svg

I understand how the theory that the maximum/minimum value will occur when the grad of function f and the grad of function g are scalar multiples of each other. But imagine for a second that the constraint curve was a straight line and it went straight through the centre of the circle created by the level curve of the function f. In that case the maximum value would be at the centre of this level curve, but to my knowledge at no time were the gradients parallel.
So like: "Minimize [itex]f(x,y)= x^2+ y^2[/itex] subject to the constraint y= x"? The constraint can be written g(x,y)= y- x so [itex]\nabla f= 2x\vec{i}+ 2y\vec{j}[/itex] and [itex]\nabla g= -vec{i}+ vec{j}[/itex]. The Lagrange multiplier method says that the max or min will be where [itex]2x= -\lambda[/itex] and [itex]2y= \lambda[/itex].

Eliminate [itex]\lambda[/itex] by dividing one equation by the other: [itex]x/y= -1[/itex] or y= -x. That, together with y= x, the constraint, give y= x= 0. The two gradients are never parallel but if one is 0 that will also satisfy [itex]\nabla g= \lambda\nabla f[/itex].

Another way of putting this is that if the two gradients are never parallel, then the "constraint" doesn't really constrain anything. The max or min will be where the gradient of the given function is 0.

Or imagine a sloping flat surface sloping up. If the constraint curve started off running up the slope then for a second ran tangent to one of the level curves and then ran at some angle up this surface and stopped, then the point at which the gradients were scalar multiples only in the middle of the curves, so the supposed maximum/minimum occurred at a random place. If you could clarify this it would be extremely helpful thanks!
 

1. What is the purpose of using Lagrangian multipliers to find maximum and minimum values?

Lagrangian multipliers are a mathematical technique used to find the maximum or minimum values of a function subject to certain constraints. This method is often used in optimization problems where we need to find the optimal value of a function while satisfying a set of constraints.

2. How do Lagrangian multipliers work?

Lagrangian multipliers work by transforming a constrained optimization problem into an unconstrained one. This is done by introducing a new variable, known as a Lagrangian multiplier, into the original function. The optimal value of this new variable can then be used to find the maximum or minimum value of the original function while satisfying the given constraints.

3. When should Lagrangian multipliers be used?

Lagrangian multipliers should be used when solving optimization problems with constraints. This method is particularly useful when the constraints are not explicitly stated in the function, making it difficult to solve using traditional optimization techniques.

4. What are the key steps in using Lagrangian multipliers to find maximum and minimum values?

The key steps in using Lagrangian multipliers are:

  • Formulating the constrained optimization problem into an unconstrained one by introducing a Lagrangian multiplier
  • Taking the partial derivatives of the new function with respect to all variables, including the Lagrangian multiplier
  • Setting the equations equal to zero and solving for the variables
  • Using the optimal value of the Lagrangian multiplier to find the maximum or minimum value of the original function

5. Can Lagrangian multipliers be used for non-linear functions?

Yes, Lagrangian multipliers can be used for both linear and non-linear functions. However, the process of solving for the optimal value of the Lagrangian multiplier may be more complex for non-linear functions compared to linear ones.

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