Solve Lagrange Multipliers Problem with x-4y=1 Constraint

In summary, the conversation discusses using Lagrange Multipliers to find the minimum and maximum values of the function f(x,y)=xy^2 under the constraint x-4y=1, with the additional constraint -1≤x≤2. The speaker is unsure how to handle the second constraint and asks for clarification. The expert summarizes the process of using Lagrange Multipliers to find the critical points and evaluates the objective function to determine the minimum and maximum values.
  • #1
Yankel
395
0
Hello all

I have this problem:

Use Lagrange Multipliers to find the min and max of:

\[f(x,y)=xy^{2}\]

under the constraint:

\[x-4y=1\]

\[-1\leqslant x\leq 2\]

My problem is: I know how to solve if

\[-1\leqslant x\leq 2\]

wasn't given. I calculate the Lagrangian function, find it's derivatives by x,y and lambda, and solve the 3 equations to find all suspicious points, I calculate the function value of them all and see which is smallest and which is largest.

I don't know what I should do with the second constraint:

\[-1\leqslant x\leq 2\]

should I simply verify that each point satisfy this condition, or should I check some boundary points ?
 
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  • #2
What I would do here is identify the objective function:

\(\displaystyle f(x,y)=xy^2\)

and the constraint:

\(\displaystyle g(x,y)=x-4y-1=0\)

And using Lagrange we ultimately find the critical points:

\(\displaystyle (x,y)=(1,0),\,\left(\frac{1}{3},-\frac{1}{6}\right)\)

Both points have $x$-values within the second constraint. Now, using the $x$-boundaries of the second constraint, we obtain two more critical points:

\(\displaystyle (x,y)=\left(-1,-\frac{1}{2}\right),\,\left(2,\frac{1}{4}\right)\)

So, you have 4 critical points at which to evaluate the objective function, and the smallest and largest values are your minimum and maximum respectively. What do you find?

Doing this, I get values that agree with W|A (if you account for approximations being done by W|A):

optimize x*y^2 subject to x-4y=1,-1<=x<=2
 

FAQ: Solve Lagrange Multipliers Problem with x-4y=1 Constraint

What is the Lagrange multiplier method?

The Lagrange multiplier method is a mathematical optimization technique used to find the maximum or minimum value of a function subject to one or more constraints. It involves finding the stationary points of the Lagrangian function, which is a combination of the original objective function and the constraints.

How does the Lagrange multiplier method work?

The Lagrange multiplier method works by introducing a new variable, called the Lagrange multiplier, for each constraint in the original problem. These multipliers act as weights, allowing the optimization problem to be solved by finding the stationary points of the Lagrangian function.

What is the purpose of using Lagrange multipliers?

The use of Lagrange multipliers allows for the optimization of a function subject to constraints, which cannot be easily solved using traditional methods. It is particularly useful in cases where the constraints are non-linear or when there are multiple constraints.

How do you solve a Lagrange multiplier problem with a linear constraint?

To solve a Lagrange multiplier problem with a linear constraint, you first need to express the constraint in the form of an equation. In this case, the constraint is x-4y=1. Next, you construct the Lagrangian function by multiplying the constraint by the Lagrange multiplier and adding it to the objective function. Then, you find the stationary points of the Lagrangian, which will give you the optimal solution.

Can Lagrange multipliers be used for non-linear constraints?

Yes, the Lagrange multiplier method can be used for non-linear constraints. However, the process becomes more complex as it involves solving a system of equations to find the stationary points of the Lagrangian function. In some cases, it may not be possible to find an analytical solution, and numerical methods may be needed.

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