Solve Lagrange Multiplier Problem | f(X,Y,Z) = 2XY + 6YZ + 8XZ

[Baumer8993]

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?

 

[Ray Vickson]

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.

 

[HallsofIvy]

So you have $$2Y+ 8Z= \lambda YZ$$, $$2X+ 6Z= \lambda XZ$$, [tex]6Y+ 8X= \lambda XY, together with XYZ= 12.<br /> <br /> Since a specific value for $$\lambda$$ is not necessary to solve this problem, it is often simplest to first eliminate $$\lambda$$ by <b>dividing</b> one equation by another. Here, dividing the first equation by the second, $$\frac{2Y+ 8Z}{2X+ 6Z}= \frac{Y}{X}$$ which is the same as X(Y+ 4Z)= Y(X+ 3Z) or XY+ 4XZ= XY+ 3YZ which reduces to 4XZ= 3YZ, 4X= 3Y.<br /> <br /> Similarly, dividing the third equation by the second, $$\frac{6Y+ 8Z}{2X+ 6Z}= \frac{Y}{Z}$$ which is the same as Z(3Y+ 4Z)= Y(X+ 3Z) or $$3YZ+ 4Z^2= XY+ 3YZ$$ which reduces to $$4Z^2= XY$$. <br /> <br /> Since Y= (4/3)X, that last equation becomes $$4Z^2= (4/3)X^2$$ or $$Z^2= 4X^2$$ and $$Z= \pm 2X$$. Continue from there.<br /> <br /> I agree with Ray Vickson- the last thing you want is to have someone or something do those solutions for you. You need to <b>learn</b> the ideas so that you will understand, when you do use "technology", what is going on.[/tex]