Lagrange Multipliers with Multiple Constraints?

[AimlessWander]

Homework Statement

Using Lagrange multipliers, find the max and the min values of f:

f(x,y,z) = x^2 +2y^2+3x^2

Constraints:

x + y + z =1
x - y + 2z = 2

Homework Equations

∇f(x) = λ∇g(x) + μ∇h(x)

The Attempt at a Solution

Using Lagrange multipliers, I obtained the equations:

2x = λ + μ
4y = λ - μ
6z = λ + 2μ

With the constraints, I tried to find values for the unknowns, but I can't solve the system of equations. I don't know where to go from here. Can anyone help put? Any input is appreciated!

 

[Ray Vickson]

Homework Statement

Using Lagrange multipliers, find the max and the min values of f:

f(x,y,z) = x^2 +2y^2+3x^2

Constraints:

x + y + z =1
x - y + 2x = 2

Homework Equations

∇f(x) = λ∇g(x) + μ∇h(x)

The Attempt at a Solution

Using Lagrange multipliers, I obtained the equations:

2x = λ + μ
4y = λ - μ
6z = λ + 2μ

With the constraints, I tried to find values for the unknowns, but I can't solve the system of equations. I don't know where to go from here. Can anyone help put? Any input is appreciated!

Your equations give you x, y and z in terms of λ and μ. Plug these expressions into the two constraint equations to get two equations for the two unknowns λ, μ.

 

[AimlessWander]

Hey, thanks for helping out! Please note the small change in the second constraint equation: it's x -y + 2z = 2. I did exactly that, but it didn't really seem to help solve for the unknowns. By plugging in the expressions obtained into the constraint equations, I got 11λ + 7μ = 12 and 7λ+ 17μ = 24. Solving, I still get an equation with two unknowns.

 

[Ray Vickson]

Hey, thanks for helping out! Please note the small change in the second constraint equation: it's x -y + 2z = 2. I did exactly that, but it didn't really seem to help solve for the unknowns. By plugging in the expressions obtained into the constraint equations, I got 11λ + 7μ = 12 and 7λ+ 17μ = 24. Solving, I still get an equation with two unknowns.

I don't understand what you mean. When you solve the two equations you will get two numbers, one for λ and one for μ. If that is not what you get you had better show the details here.

BTW: you wrote f(x,y,z) = x^2 +2y^2+3x^2, but I assume you meant 3z^2 in the last term.

 

[AlephZero]

Altogether you have 5 equations in 5 unknowns x, y, z, λ, and μ: the two constraint equations, and the three equations you have derived.

There are lots of ways to solve them (some quicker than others) but you certainly have enough equations to get the solution.

You don't need the values of λ and μ to get the max and min values of f, but as Ray Vickson said it's probably easiest to find λ and μ first, and then find x y and z.

 

[HallsofIvy]

With 2x= \lambda+ \mu, 2y= \lambda-\mu, and 2z= \lambda+ 2\mu the two constraints become x+ y+ z= (\lambda + \mu)/2+ (\lambda- \mu)/2+ (\lambda+ 2\mu)/2= 1, which reduces to 3\lambda+ 2\mu= 2, and x- y+ 2z= (\lambda+ \mu)/2- (\lambda- \mu)/2+ 2(\lambda+ 2\mu)/2= 2, which reduces to \lambda+ 2\mu= 2.

Solve 3\lambda+ 2\mu= 2 and \lambda+ 3\mu= 2.