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Lagrange Multiplier

  • Thread starter Baumer8993
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  • #1
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Homework Statement



Minimize f(X, Y, Z) = 2XY + 6YZ + 8XZ subject to the constraint XYZ = 12.


Homework Equations


The gradients of the equations, and XYZ = 12.


The Attempt at a Solution



I have the gradients for both of the equations.

∇f = <2Y + 8Z, 2X + 6Z, 6Y + 8X> ∇g = < YZ, XZ, XY> I also have XYZ 12.

I have set them up in there separate equations with the λ in them. I have just stuck on how to solve them. Isn't there a way to plug the equations in the my TI - 83 Plus to solve them?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Minimize f(X, Y, Z) = 2XY + 6YZ + 8XZ subject to the constraint XYZ = 12.


Homework Equations


The gradients of the equations, and XYZ = 12.


The Attempt at a Solution



I have the gradients for both of the equations.

∇f = <2Y + 8Z, 2X + 6Z, 6Y + 8X> ∇g = < YZ, XZ, XY> I also have XYZ 12.

I have set them up in there separate equations with the λ in them. I have just stuck on how to solve them. Isn't there a way to plug the equations in the my TI - 83 Plus to solve them?
No, don't do that; you won't learn anything that way. Use such tools later, to save time, after you have learned the basics.

In this case: use the first equation to solve for Y in terms of Z, use the second equation to solve for X in terms of Z. Now plug those tow expressions into the third equation. This will give you an equation in Z and λ alone. One way to proceed would be to try to solve that equation for Z in terms of λ; there would be more than one solution, corresponding to multiple roots of a higher-order equation. Another way would be to also substitute your expressions for X and Y into the constraint equation, giving a second equation involving z and λ. Now you could try to solve those two equations in the two unknowns in some way.
 
  • #3
HallsofIvy
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So you have [tex]2Y+ 8Z= \lambda YZ[/tex], [tex]2X+ 6Z= \lambda XZ[/tex], [tex]6Y+ 8X= \lambda XY, together with XYZ= 12.

Since a specific value for [tex]\lambda[/tex] is not necessary to solve this problem, it is often simplest to first eliminate [tex]\lambda[/tex] by dividing one equation by another. Here, dividing the first eqution by the second, [tex]\frac{2Y+ 8Z}{2X+ 6Z}= \frac{Y}{X}[/tex] which is the same as X(Y+ 4Z)= Y(X+ 3Z) or XY+ 4XZ= XY+ 3YZ which reduces to 4XZ= 3YZ, 4X= 3Y.

Similarly, dividing the third equation by the second, [tex]\frac{6Y+ 8Z}{2X+ 6Z}= \frac{Y}{Z}[/tex] which is the same as Z(3Y+ 4Z)= Y(X+ 3Z) or [tex]3YZ+ 4Z^2= XY+ 3YZ[/tex] which reduces to [tex]4Z^2= XY[/tex].

Since Y= (4/3)X, that last equation becomes [tex]4Z^2= (4/3)X^2[/tex] or [tex]Z^2= 4X^2[/tex] and [tex]Z= \pm 2X[/tex]. Continue from there.

I agree with Ray Vickson- the last thing you want is to have someone or something do those solutions for you. You need to learn the ideas so that you will understand, when you do use "technology", what is going on.
 
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