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Solving an equation with variable as denominator?

  • Thread starter CptDarling
  • Start date
  • #1

Homework Statement


The problem initially was a Lagrange multipliers question, and I'm trying to solve for λ. I have an equation in terms of λ alone, but I just can't solve it. I plugged it into Wolfram Alpha and it gave me the λ value in the back of the textbook, but I still don't know how to solve the equation myself.


Homework Equations


4(1/2λ)^2-λ=0


The Attempt at a Solution



4(1^2/2λ^2)-λ=0
4/2λ^2=λ
2/λ^2=λ

And I don't know where to go from here
 

Answers and Replies

  • #2
HallsofIvy
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By [itex]4(1/2\lambda)^2= \lambda[/itex], I presume you mean [itex]4(1/(2\lambda)^2= \lambda[/itex] rather than [itex]4(1/2)^2\lambda)^2= \lambda[/itex]

If that is correct, an obvious first step would be to cancel the "4"s to get
[tex]\frac{1}{\lambda^2}= \lambda[/tex]
and the multiply on both sides by [itex]\lambda[/itex].

But I am puzzled as to why you would want to find [itex]\lambda[/itex]. It is not part of the solution to a Lagrange multipliers problem and it is never necessary to find it to solve a Lagrange multipliers problem.

Using Lagrange multipliers always gives a series of equation like:
[itex]f_1(x,y,z)= \lambda g_1(x,y,z)[/itex]
[itex]f_2(x,y,z)= \lambda g_2(x,y,z)[/itex]
[itex]f_3(x,y,z)= \lambda g_3(x,y,z)[/itex]

and a good first step is to eliminate [itex]\lambda[/itex] by dividing on equation by another.
 
  • #3
Ray Vickson
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By [itex]4(1/2\lambda)^2= \lambda[/itex], I presume you mean [itex]4(1/(2\lambda)^2= \lambda[/itex] rather than [itex]4(1/2)^2\lambda)^2= \lambda[/itex]

If that is correct, an obvious first step would be to cancel the "4"s to get
[tex]\frac{1}{\lambda^2}= \lambda[/tex]
and the multiply on both sides by [itex]\lambda[/itex].

But I am puzzled as to why you would want to find [itex]\lambda[/itex]. It is not part of the solution to a Lagrange multipliers problem and it is never necessary to find it to solve a Lagrange multipliers problem.

Using Lagrange multipliers always gives a series of equation like:
[itex]f_1(x,y,z)= \lambda g_1(x,y,z)[/itex]
[itex]f_2(x,y,z)= \lambda g_2(x,y,z)[/itex]
[itex]f_3(x,y,z)= \lambda g_3(x,y,z)[/itex]

and a good first step is to eliminate [itex]\lambda[/itex] by dividing on equation by another.
Actually, finding λ is often very important in solving a constrained optimization problem. Often, for example, we find x(λ), y(λ) and z(λ) from the optimality conditions, then find itself (and thus the values of x, y and z) by solving an equation in λ obtained from the constraint. Many optimization codes and methods for numerical solution depend crucially on estimating Lagrange multipliers along with the variables, and many of the very best methods would not work at all without doing this.

RGV
 
  • #4
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By [itex]4(1/2\lambda)^2= \lambda[/itex], I presume you mean [itex]4(1/(2\lambda)^2= \lambda[/itex] rather than [itex]4(1/2)^2\lambda)^2= \lambda[/itex]
I doubt it. You really are very sloppy, Ivy! :smile:
 

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