Thread: Using Lagrange Multiploiers locate any extrema of the function

1. Using Lagrange Multiploiers locate any extrema of the function

Using Lagrange Multiploiers locate any extrema of the function
x^2 + y^2 + z^2
subject to x + z = 6 and y + z = 8

I got (4/3, 10/3, 14/3)

2. No you didn't. Check your arithmetic. You are SO CLOSE.

Hint: The function is symmetric, but the constraints are not.

If you're still struggling, show your work and guidance can be offered.

3. Originally Posted by TKHunny
No you didn't. Check your arithmetic. You are SO CLOSE.

Hint: The function is symmetric, but the constraints are not.

If you're still struggling, show your work and guidance can be offered.
Mmmm... I don't know about the OP, but I am having trouble interpreting this comment.

This problem can be transformed into a 1-D unconstrained optimisation problem in the variable z. This transformed problem has a unique minimum at z=14/3, which then gives the x and y for the minimum by substitution into the constraints.

CB

4. Originally Posted by TKHunny
No you didn't. Check your arithmetic. You are SO CLOSE.

Hint: The function is symmetric, but the constraints are not.

If you're still struggling, show your work and guidance can be offered.
I see nothing wrong with the arithmetic.

As Captain Black says, you can use the two constraints to reduce this to a quadratic function of the single variable, z, then show that z= 14/3 is a minimum for that function and solve for x and y from the constraints.

But since this says to use "Lagrange Multipliers", we have $2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(\vec{i}+ \vec{k})+ \mu(\vec{j}+ \vec{k})$ where $\lambda$ and $\mu$ are the multipliers.

That gives the three equations $2x= \lambda$, $2y= \mu$, and $2z= \lambda+ \mu$. Since $2x+ 2y= \lambda+ \mu$, that is the same as 2z= 2x+ 2y or z= x+ y. Then x= z- y and the first constraint is 2z- y= 6 while the second is y+ z= 8. Adding those equations 3z= 14 so z= 14/3. Then y= 8- 14/3= 10/3 and x= 4/3 .

5. Originally Posted by HallsofIvy
I see nothing wrong with the arithmetic.

As Captain Black says, you can use the two constraints to reduce this to a quadratic function of the single variable, z, then show that z= 14/3 is a minimum for that function and solve for x and y from the constraints.

But since this says to use "Lagrange Multipliers", we have $2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(\vec{i}+ \vec{k})+ \mu(\vec{j}+ \vec{k})$ where $\lambda$ and $\mu$ are the multipliers.

That gives the three equations $2x= \lambda$, $2y= \mu$, and $2z= \lambda+ \mu$. Since $2x+ 2y= \lambda+ \mu$, that is the same as 2z= 2x+ 2y or z= x+ y. Then x= z- y and the first constraint is 2z- y= 6 while the second is y+ z= 8. Adding those equations 3z= 14 so z= 14/3. Then y= 8- 14/3= 10/3 and x= 4/3 .
That is not a notation for Lagrange multiplers that I recognise, but my Lagrange multipler result is the same. I only brought up the unconstrained method as a cross check that there was not something going on that I had missed in the Largangian method.

CB

6. That's odd. I could have sworn the 14 and the 10 were reversed. Has it been edited or was I seeing cross-eyed last night? No matter.