Using Lagrange Multiploiers locate any extrema of the function

**ewkimchi** · June 2nd 2010, 08:57 AM

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)

**TKHunny** · June 2nd 2010, 06:42 PM

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.

**CaptainBlack** · June 2nd 2010, 11:34 PM

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

**HallsofIvy** · June 3rd 2010, 02:37 AM

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 .

**CaptainBlack** · June 3rd 2010, 02:44 AM

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

**TKHunny** · June 3rd 2010, 09:57 AM

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.

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