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

Click For Summary
SUMMARY

The discussion focuses on finding the minimum and maximum values of the function f(x,y,z) = 3x + 2y + 4z, constrained by the equation x² + 2y² + 6z² = 81. Participants shared their approaches using the method of Lagrange multipliers, setting up equations based on the gradients of the function and constraint. Key equations derived include 3 = 2x(λ), 2 = 4y(λ), and 4 = 12z(λ), leading to relationships between x, y, and z. The final step involves substituting expressions for y and z in terms of x back into the constraint to solve for x.

PREREQUISITES
  • Understanding of multivariable calculus, specifically Lagrange multipliers.
  • Familiarity with gradient vectors and their applications in optimization problems.
  • Knowledge of quadratic forms and constraints in optimization.
  • Ability to manipulate algebraic expressions and solve equations.
NEXT STEPS
  • Study the method of Lagrange multipliers in detail, including examples and applications.
  • Learn how to derive and solve systems of equations resulting from optimization problems.
  • Explore quadratic forms and their significance in constrained optimization.
  • Practice solving similar optimization problems with different constraints and objective functions.
USEFUL FOR

Students and professionals in mathematics, engineering, and economics who are involved in optimization problems, particularly those using Lagrange multipliers for constrained optimization.

ajkess1994
Messages
9
Reaction score
0
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)
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
6K
  • · Replies 46 ·
2
Replies
46
Views
7K
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K