Find the minimum of (4xyz)/(3)+x²+y²+z²

  • Context: MHB 
  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Minimum
Click For Summary
SUMMARY

The discussion focuses on minimizing the expression \(\frac{4xyz}{3} + x^2 + y^2 + z^2\) under the constraint \(x + y + z = 3\), where \(x\), \(y\), and \(z\) are the lengths of the sides of a triangle. MarkFL provided a solution that effectively addresses the problem, demonstrating a clear understanding of the relationship between the variables and the constraints imposed by the triangle inequality. The conclusion emphasizes the importance of both algebraic manipulation and geometric interpretation in solving optimization problems involving triangle side lengths.

PREREQUISITES
  • Understanding of triangle inequalities and properties
  • Familiarity with optimization techniques in calculus
  • Knowledge of algebraic manipulation and expressions
  • Basic understanding of Lagrange multipliers for constrained optimization
NEXT STEPS
  • Study the application of Lagrange multipliers in optimization problems
  • Explore geometric interpretations of algebraic expressions
  • Learn about the AM-GM inequality and its applications in optimization
  • Investigate other optimization problems involving constraints on variables
USEFUL FOR

Mathematicians, students studying optimization techniques, and anyone interested in solving constrained optimization problems in geometry.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $x,\,y,\,z$ be the lengths of the sides of a triangle such that $x+y+z=3$.

Find the minimum of $\dfrac{4xyz}{3}+x^2+y^2+z^2$.
 
Physics news on Phys.org
My solution:

We see that there is cyclic symmetry in the objective function $f$ and constraint $g$, thus the critical point is:

$$(x,y,z)=(1,1,1)$$

We then find:

$$f(1,1,1)=\frac{13}{3}$$

Testing another point:

$$f\left(\frac{1}{2},\frac{1}{2},2\right)>\frac{13}{3}$$

Hence:

$$f_{\min}=\frac{13}{3}$$
 
MarkFL said:
My solution:

We see that there is cyclic symmetry in the objective function $f$ and constraint $g$, thus the critical point is:

$$(x,y,z)=(1,1,1)$$

We then find:

$$f(1,1,1)=\frac{13}{3}$$

Testing another point:

$$f\left(\frac{1}{2},\frac{1}{2},2\right)>\frac{13}{3}$$

Hence:

$$f_{\min}=\frac{13}{3}$$

Very well done, MarkFL! (Sun) And thanks for participating!
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
912
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K