Saladsamurai said:Homework Statement
Find the critical points of [itex]f(x,y)=x+y^2[/itex] subject to the constraint [itex]g(x)=x^2+y^2=1[/itex]
Homework Equations
[itex]\nabla f=\lambda\nabla g[/itex]
[itex]g(x,y)=1[/itex]
The Attempt at a Solution
[tex]f_x=1=2\lambda*x\Rightarrow x=\frac{1}{2\lambda}[/tex]
[tex]f_y=2y=2\lambda*y\Rightarrow y(\lambda-1)=0\Rightarrow y=0 \\ or\\ \lambda=1[/tex]
[tex]x^2+y^2=1[/tex]
I am a little confused as to where I go from here?
LeGrange multipliers are a mathematical tool used to find critical points of a function subject to one or more constraints. They are used in optimization problems to find the maximum or minimum values of a function with certain restrictions.
To find the critical points of a function using LeGrange multipliers, you first need to set up the Lagrangian function by combining the original function with the constraints using a Lagrange multiplier. Then, you take the partial derivatives of the Lagrangian function with respect to each variable, set them equal to zero, and solve for the variables. The resulting values are the critical points of the function.
LeGrange multipliers are commonly used in economics, engineering, physics, and other fields to solve optimization problems. They can be used to maximize profits, minimize costs, determine the optimal design of a structure, and much more.
Yes, LeGrange multipliers can be extended to functions with multiple variables and constraints. In these cases, the Lagrangian function becomes more complex, but the process for finding the critical points remains the same.
One limitation of LeGrange multipliers is that they may not always yield the global maximum or minimum of a function. In some cases, they may only find a local maximum or minimum. Additionally, the process for finding the Lagrangian function and solving for the critical points can be time-consuming and complex for certain functions.