Using Lagrange multipliers w/ f(x,y,z)=x^2-3y^2-z^2+6 subj. to 5x-3y+z=21

[RandyH]

7) Use the method of Lagrange multipliers to determine the critical points of \(\displaystyle f(x,\,y,\,z)\, =\, x^2\, -\, 3y^2\, -\, z^2\, +\, 6\) subject to the constraint \(\displaystyle 5x\, -\, 3y\, +\, z\, =\, 21\)

https://www.freemathhelp.com/forum/...sapp.com/6e721586-c904-4850-85e4-3e0c847543f4

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This one question seems a lot more difficult than using Lagrange when it's just f(x,y). I set up the Lagrange so it looked like this:

L = x^2 - 3y^2 - z^2 + 6 + Lambda(21 - 5x + 3y - z)

Then I differentiate with respect to each variable to get:

L_x = 2x - 5*Lambda = 0
L_y = -6y + 3*Lambda = 0
L_z = 2z - Lambda = 0
L_Lambda = 21 - 5x - 3y - z = 0

From my POV this is where things start to go wrong. I have 4 equations so I should be able to solve for Lambda, X, Y, and Z? I tried using substitution and elimination, but I keep going in circles and end up with a wrong answer.

Thanks for any help in advance.

 

[Deleted member 4993]

View attachment 9504

This one question seems a lot more difficult than using Lagrange when it's just f(x,y). I set up the Lagrange so it looked like this:

L = x^2 - 3y^2 - z^2 + 6 + Lambda(21 - 5x + 3y - z)

Then I differentiate with respect to each variable to get:

L_x = 2x - 5*Lambda = 0
L_y = -6y + 3*Lambda = 0
L_z = 2z - Lambda = 0
L_Lambda = 21 - 5x - 3y - z = 0

From my POV this is where things start to go wrong. I have 4 equations so I should be able to solve for Lambda, X, Y, and Z? I tried using substitution and elimination, but I keep going in circles and end up with a wrong answer.

Thanks for any help in advance.

Using MS-excel and matrix inversion, I get:

x =	3.62069
y =	0.724138
z =	0.724138
l =	1.448276

So x = 5y = 5z = 2.5 l

 

[HallsofIvy]

I would write those equations involving \(\displaystyle \lambda\) as
\(\displaystyle 2x= 5\lambda\)
\(\displaystyle -6y= -3\lambda\) and
\(\displaystyle 2z= \lambda\).


Since a specific value for \(\displaystyle \lambda\) is not necessary for a solution often a good first step is to eliminate \(\displaystyle \lambda\) by dividing one equation by another.

Dividing the first equation by the second gives
\(\displaystyle -\frac{x}{3y}= -\frac{5}{3}\) which is equivalent to x= 5y.

Dividing the second equation by the third gives
\(\displaystyle -\frac{3y}{z}= -3\) which is equivalent to y= z.

So the constraint becomes 5x- 3y+ z= 5(5z)- 3z+ z= 23z= 21.
z= 21/23

y= z= 21/23

x= 5y= 105/23.

 

[JeffM]

View attachment 9504

This one question seems a lot more difficult than using Lagrange when it's just f(x,y). I set up the Lagrange so it looked like this:

L = x^2 - 3y^2 - z^2 + 6 + Lambda(21 - 5x + 3y - z)

Then I differentiate with respect to each variable to get:

L_x = 2x - 5*Lambda = 0
L_y = -6y + 3*Lambda = 0
L_z = 2z - Lambda = 0
L_Lambda = 21 - 5x - 3y - z = 0

From my POV this is where things start to go wrong. I have 4 equations so I should be able to solve for Lambda, X, Y, and Z? I tried using substitution and elimination, but I keep going in circles and end up with a wrong answer.

Thanks for any help in advance.

The method you proposed works and is quite straight forward. We cannot tell you where you went wrong because you did not show your work, but it is basic algebra.

\(\displaystyle 2z - \lambda = 0 \implies \lambda = 2z.\)

\(\displaystyle -\ 6y + 3 \lambda = 0 \implies 2y = \lambda = 2z \implies y = z.\)

\(\displaystyle 2x - 5 \lambda = 0 \implies 2x = 5 \lambda = 5 * 2y \implies x = 5y = 5z.\)

\(\displaystyle 0 = 21 - 5x + 3y - z = 21 - 5(5y) + 3y - y = 21 - 23y \implies\)

\(\displaystyle y = \dfrac{21}{23} = z \implies x = \dfrac{105}{23}.\)

As I said, just some simple algebra.