How Do You Apply Lagrange Multipliers to Find Extremes on a Circle and Disc?

In summary, the maximum and minimum values for the function f(x,y) = x5y3 on the circle defined by x2 + y2 = 10 can be found by minimizing f subject to g(x,y) = 0. For the function f(x,y) = x5y3 on the disc x2 + y2 ≤ 10, the maximum and minimum values can be found by maximizing f subject to g(x,y) = 0. The '10' is included in the definition of g for the first problem, but is not included in the definition of g for the second problem.
  • #1
Locoism
81
0

Homework Statement



Find the maximum and minimum values of f(x,y) = x5y3 on the circle defined by x2 + y2 = 10. Do the same for the disc x2 + y2 ≤ 10.


The Attempt at a Solution


for the first part, if I call the circle g(x,y) defined by x2 + y2 = 10
I need to now define some F(x,y,λ) = f(x,y) - λg(x,y) and find the critical points of this, i.e.: where Fx', Fy', and Fλ' = 0. I was wondering if I should include the constant (10) in the function λg, or does it matter at all? i.e.:
F(x,y,λ) = x5y3 - λx2 - λy2 + 10λ
OR
F(x,y,λ) = x5y3 - λx2 - λy2

(from there I can figure it out) I was just unclear as to how the constraint is handled.
 
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  • #2
Locoism said:

Homework Statement



Find the maximum and minimum values of f(x,y) = x5y3 on the circle defined by x2 + y2 = 10. Do the same for the disc x2 + y2 ≤ 10.


The Attempt at a Solution


for the first part, if I call the circle g(x,y) defined by x2 + y2 = 10
I need to now define some F(x,y,λ) = f(x,y) - λg(x,y) and find the critical points of this, i.e.: where Fx', Fy', and Fλ' = 0. I was wondering if I should include the constant (10) in the function λg, or does it matter at all? i.e.:
F(x,y,λ) = x5y3 - λx2 - λy2 + 10λ
OR
F(x,y,λ) = x5y3 - λx2 - λy2

(from there I can figure it out) I was just unclear as to how the constraint is handled.

I doubt think the constant makes a difference as you'll have to differentiate g,f.
 
  • #3
yes but it matters when you differentiate for λ
 
  • #4
Differentiating with respect to [itex]\lambda[/itex] gives you the equation you already know [itex]x^2 + y^2 = 10[/itex]. So for the sake of completeness, you should keep the 10 in the equation.

But if you're not interested in differentiating with respect to [itex]\lambda[/itex] (because you already know what it will tell you), then you could just take the 10 out.
 
  • #5
alright thank you!
 
  • #6
Locoism said:

Homework Statement



Find the maximum and minimum values of f(x,y) = x5y3 on the circle defined by x2 + y2 = 10. Do the same for the disc x2 + y2 ≤ 10.


The Attempt at a Solution


for the first part, if I call the circle g(x,y) defined by x2 + y2 = 10
I need to now define some F(x,y,λ) = f(x,y) - λg(x,y) and find the critical points of this, i.e.: where Fx', Fy', and Fλ' = 0. I was wondering if I should include the constant (10) in the function λg, or does it matter at all? i.e.:
F(x,y,λ) = x5y3 - λx2 - λy2 + 10λ
OR
F(x,y,λ) = x5y3 - λx2 - λy2

(from there I can figure it out) I was just unclear as to how the constraint is handled.

The _standard_ Lagrangian approach to max/min f(x1,x2,...,xn) subject to g(x1,x2,..,xn) = 0 is to look at L = f +- u*g (using u instead of lambda) and to set the partials dL/d x_j = 0 for j = 1,...,n. So, in your case, g = x^2 + y^2 - 10; that is, the '10' is part of the definition of g. Of course it goes away when you take xj-derivatives. Keeping the '10' is certainly necessary if you set dL/du = 0.

Why does this matter? Well, if you use the convention L = f - u*g, and look at a 'nearby' problem max/min f subject to g = a (for small |a|) then the optimal value function F(a) (= optimal value of f, subject to g(x) = a) satisfies dF/da|_{a=0} = u, where u = the Lagrange multiplier in the a=0 problem. So, for example, suppose you wanted the maximum value (not the x,y that give the maximum) for the related problem x^2 + y^2 = 10.1. This corresponds to g = 0.1, because the '10' is part of the definition of g. The optimal value in the new problem would be (approx) the old optimal value + 0.1*u .

RGV
 
  • #7
But for the second part, the disk defined by x2 + y2 ≤ 10,
How is this any different? at least in this case, the max and min are on the outer circle so there is no max or min when x and y are less. Is there a different method of solving this?
 
  • #8
I believe for the second part all you have to show is that the max occurs on the boundary and so the answer should be the same as part 1. Extremize f, and show that there is only a minimum at some point inside the boundary, and then use the extreme-value theorem to show that the maximum must be on the boundary (i.e. a maximum must exist).
 
  • #9
Locoism said:
But for the second part, the disk defined by x2 + y2 ≤ 10,
How is this any different? at least in this case, the max and min are on the outer circle so there is no max or min when x and y are less. Is there a different method of solving this?

You need to worry about whether or not there is an interior optimum (where x^2 + y^2 < 10); at such a point the Lagrange multiplier = 0 and you are just looking at stationary points of f(x,y) itself. In this case a direct argument is easiest: given any interior point (x0,y0), argue that there are other points (x1,y1) and (x2,y2) giving f(x1,x1) > f(x0,y0) > f(x2,y2); you actually need to carry out that argument, but it is not hard.

Note: in this case it is important that x and y not have sign restrictions (so negative x and/or y are OK). If you also had restrictions x >= 0 and y >= 0, you would need to be more careful.

RGV
 

Related to How Do You Apply Lagrange Multipliers to Find Extremes on a Circle and Disc?

What is a Lagrange multiplier question?

A Lagrange multiplier question is a type of optimization problem in mathematics, where the objective is to maximize or minimize a function subject to one or more constraints.

How is a Lagrange multiplier used to solve a problem?

A Lagrange multiplier is a scalar value that is added to the objective function in order to incorporate the constraints into the optimization problem. This allows for the use of the method of Lagrange multipliers to find the optimal solution.

What are the steps for solving a Lagrange multiplier question?

The steps for solving a Lagrange multiplier question are:
1. Formulate the objective function and the constraints.
2. Set up the Lagrangian function by adding the Lagrange multiplier to the objective function.
3. Take the partial derivatives of the Lagrangian with respect to each variable.
4. Set the partial derivatives equal to zero and solve the resulting system of equations for the optimal solution.
5. Check the second-order conditions to ensure that the solution is a maximum or minimum.
6. Substitute the optimal solution back into the original objective function to find the maximum or minimum value.

What types of problems can be solved using Lagrange multipliers?

Lagrange multipliers can be used to solve a variety of optimization problems, including constrained optimization, economics, engineering, and physics problems.

What are some common applications of Lagrange multipliers in real life?

Lagrange multipliers have many real-life applications, such as finding the minimum cost of production for a company, optimizing resource allocation in economics, determining the optimal trajectory for a spacecraft, and minimizing energy usage in engineering projects.

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