Solve f(x,y,z): Min & Max Values Subject to Constraint x^2+2y^2+6z^2=81

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Discussion Overview

The discussion revolves around finding the minimum and maximum values of the function f(x,y,z)=3x+2y+4z under the constraint x^2+2y^2+6z^2=81. Participants explore methods for solving this optimization problem, including the use of Lagrange multipliers and substitution techniques.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant expresses difficulty in solving the problem, noting that their equations lead to trivial identities (x=x, y=y, z=z).
  • Another participant suggests expressing y and z in terms of x and substituting these into the constraint equation.
  • A further contribution proposes specific substitutions for y and z based on the relationships derived from the Lagrange multiplier equations.
  • There is a confirmation from one participant that the suggested process worked for them.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using substitutions to simplify the problem, but there is no consensus on the final values for fmax and fmin, as these have not been explicitly calculated or confirmed in the discussion.

Contextual Notes

Participants have not resolved the mathematical steps to find the exact minimum and maximum values, and the discussion does not clarify the assumptions made during the substitutions.

ajkess1994
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Afternoon, I have been working on this problem for awhile now but have been stuck on a certain point, and once I set everything equal to each other I end up with the same thing for example: x=x, y=y, & z=z

Find the minimum and maximum values of the function f(x,y,z)=3x+2y+4z subject to the constraint x^(2)+2y^(2)+6z^(2)=81.

fmax= ?
fmin= ?

This is what I have,

fx=3, fy=2, fz=4, gx=2x, gy=4y, gz=12z

3=2x(lamda); (lamda)= 3/2x

2=4y(lamda); (lamda)= 2/4y

4=12y(lamda); (lamda)= 4/12z

(3/2x)=(2/4y)=(4/12z)
 
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ajkess1994 said:
Afternoon, I have been working on this problem for awhile now but have been stuck on a certain point, and once I set everything equal to each other I end up with the same thing for example: x=x, y=y, & z=z

Find the minimum and maximum values of the function f(x,y,z)=3x+2y+4z subject to the constraint x^(2)+2y^(2)+6z^(2)=81.

fmax= ?
fmin= ?

This is what I have,

fx=3, fy=2, fz=4, gx=2x, gy=4y, gz=12z

3=2x(lamda); (lamda)= 3/2x

2=4y(lamda); (lamda)= 2/4y

4=12y(lamda); (lamda)= 4/12z

(3/2x)=(2/4y)=(4/12z)

Hi ajkess1994 and welcome to MHB! ;)

Express $y$ in $x$, and also $z$ in $x$, and substitute in $x^2+2y^2+6z^2=81$?
 
ajkess1994 said:
Afternoon, I have been working on this problem for awhile now but have been stuck on a certain point, and once I set everything equal to each other I end up with the same thing for example: x=x, y=y, & z=z

Find the minimum and maximum values of the function f(x,y,z)=3x+2y+4z subject to the constraint x^(2)+2y^(2)+6z^(2)=81.

fmax= ?
fmin= ?

This is what I have,

fx=3, fy=2, fz=4, gx=2x, gy=4y, gz=12z

3=2x(lamda); (lamda)= 3/2x

2=4y(lamda); (lamda)= 2/4y

4=12y(lamda); (lamda)= 4/12z

(3/2x)=(2/4y)=(4/12z)

I would first look at:

$$\frac{3}{2x}=\frac{2}{4y}\implies y=\frac{x}{3}$$

And then:

$$\frac{3}{2x}=\frac{4}{12z}\implies z=\frac{2x}{9}$$

Now, substitute into your constraint, and solve for \(x\)...what do you get?
 
Thank you MarkFL the process carried out and worked.
 

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