Find Relative Extrema of f|s: Explained with Lagrange Multipliers

In summary, the problem is to find the relative extrema of a function f|S, which is a subset of R3. The approach would be to find the extreme points of f within S and solve for all variables.
  • #1
RaulTheUCSCSlug
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I am in Calculus 3, and I do not under stand what it means when they ask to find the relative extrema of f|S?

The problem is usually something like f:R^n=>R, (x,y,z) |=> (some function) , S= {(x,y) | x e R}

What does f|s mean? How does this relate to Lagrange multipliers? The book does not have an example of this notation and I can not seem to find any videos using this notation.
 
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  • #2
RaulTheUCSCSlug said:
I am in Calculus 3, and I do not under stand what it means when they ask to find the relative extrema of f|S?

The problem is usually something like f:R^n=>R, (x,y,z) |=> (some function) , S= {(x,y) | x e R}

What does f|s mean? How does this relate to Lagrange multipliers? The book does not have an example of this notation and I can not seem to find any videos using this notation.
I haven't seen that notation, but my guess is that f|S means f restricted to points in S.

BTW, what you wrote doesn't make much sense. If f is a map from Rn to R, the domain for f would be an n-tuple, not a triple like (x, y, z). Also, in your description for set S, what you wrote says that S is a set of ordered pairs (x, y), with x a real number, but nothing is stated about y.
 
  • #3
That is why I am so confused about. I thought it was f restricted to points in S but have never heard of such notation. Would it help if I gave you an example of one of the problems from the book.
 
  • #4
RaulTheUCSCSlug said:
That is why I am so confused about. I thought it was f restricted to points in S but have never heard of such notation. Would it help if I gave you an example of one of the problems from the book.
Yes, that would be helpful.
 
  • #5
As it is written in the book:

"Find the relative extrema of f|S

ƒ: R2→R, (x,y,z) |→ x2+y2+z2, S= {(x,y,z) | z≥2+x2+y2

"
Would I solve this problem like a lagrange multiplier? As in I subtract S from ƒ and find the zeroes, then plug back into the original equation or what would be the approach? How does the greater or equal to sign effect the problem? I know it will restrict it, but wouldn't it be better to have it as greater or equal to 0 and subtract the z to the other side so that I may solve for all variables?

I am confused on the notation and approach that they want me to take, since there are several problems like this in the exercises of the book, but no examples.
 
  • #6
RaulTheUCSCSlug said:
As it is written in the book:

"Find the relative extrema of f|S

ƒ: R2→R, (x,y,z) |→ x2+y2+z2, S= {(x,y,z) | z≥2+x2+y2

"
Would I solve this problem like a lagrange multiplier? As in I subtract S from ƒ and find the zeroes, then plug back into the original equation or what would be the approach? How does the greater or equal to sign effect the problem? I know it will restrict it, but wouldn't it be better to have it as greater or equal to 0 and subtract the z to the other side so that I may solve for all variables?

I am confused on the notation and approach that they want me to take, since there are several problems like this in the exercises of the book, but no examples.
I would probably take a different approach, mostly because it's been a long time since I did anything with the Lagrange multipliers approach. Based on what you wrote, I'm pretty sure they mean for you to find the extreme for f restricted to set S.

Geometrically, S is a paraboloid that opens upward, with its low point at (0, 0, 2), plus all of the points inside the paraboloid. Any extrema would have to occur where all three partials are zero, as well as potentially on the boundary of S.

As in I subtract S from ƒ and find the zeroes
This doesn't make much sense. You can't subtract a set from a function.

You also have a typo in this:
ƒ: R2→R, (x,y,z) |→ x2+y2+z2, S= {(x,y,z) | z≥2+x2+y2
f is a map from R3 to R, so the first part of what you wrote should be ƒ: R3→R. The domain for f is a subset of R3, namely the paraboloid I mentioned.
 
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  • #7
Right, must have written it down wrong. I went to my TA session today and was able to clear up the rest of the confusion I had. But apparently f|S is a function constrained by the parameters in S. Thank you for helping!
 
  • #8
RaulTheUCSCSlug said:
But apparently f|S is a function constrained by the parameters in S. Thank you for helping!
It is f restricted to (or constrained to) the set S, not "the parameters in S" which doesn't make any sense.

IOW, we're considering only those points in R3 that happen to lie on or inside the paraboloid.
 

Related to Find Relative Extrema of f|s: Explained with Lagrange Multipliers

What is the purpose of finding relative extrema using Lagrange Multipliers?

The purpose of finding relative extrema using Lagrange Multipliers is to optimize a function subject to one or more constraints. This method allows us to find the maximum or minimum values of a multivariable function while considering the constraints.

How do Lagrange Multipliers work?

Lagrange Multipliers use the method of partial derivatives to find the critical points of a function. These critical points are then compared to the constraints to determine the relative extrema.

What are the steps involved in finding relative extrema with Lagrange Multipliers?

The steps involved in finding relative extrema with Lagrange Multipliers are:
1. Write the function to be optimized and the constraints in the form of equations.
2. Create a new function by multiplying the constraints with a constant (the Lagrange Multiplier).
3. Take the partial derivatives of this new function with respect to all variables.
4. Set the partial derivatives equal to 0 and solve the resulting system of equations.
5. Substitute the solutions into the original function to find the relative extrema.

What are some common mistakes made when using Lagrange Multipliers?

Some common mistakes made when using Lagrange Multipliers include:
1. Forgetting to include all constraints in the new function.
2. Not taking the partial derivatives correctly.
3. Solving the system of equations incorrectly.
4. Not considering all possible solutions.
5. Not checking the second-order conditions to ensure the relative extrema are maximum or minimum points.

Can Lagrange Multipliers be used for functions with more than two variables?

Yes, Lagrange Multipliers can be used for functions with any number of variables. The process is the same, but the system of equations will have more equations and variables to solve for.

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