Lagrange Multipliers - 2 questions

Click For Summary
SUMMARY

The discussion revolves around solving two optimization problems using Lagrange multipliers. The first problem involves finding the hottest and coldest points on a circular plate defined by the region D = {(x1, x2) | x1² + x2² ≤ 1} with the temperature function T = 2x² - 3y² - 2x. The second problem demonstrates that for positive real numbers (x1, x2, x3) satisfying the constraint x1x2x3 = 1, the sum x1 + x2 + x3 is at least 3. The solutions indicate that both problems require finding critical points by analyzing gradients and applying boundary conditions.

PREREQUISITES
  • Understanding of Lagrange multipliers for constrained optimization
  • Knowledge of gradient calculations in multivariable calculus
  • Familiarity with circular regions in Cartesian coordinates
  • Basic principles of inequalities and their proofs in real analysis
NEXT STEPS
  • Study the method of Lagrange multipliers in detail, focusing on applications in optimization problems
  • Learn how to compute gradients for multivariable functions
  • Explore the concept of boundary conditions in optimization problems
  • Investigate inequalities in real analysis, particularly the AM-GM inequality
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, optimization, and real analysis. This discussion is beneficial for anyone looking to deepen their understanding of Lagrange multipliers and optimization techniques.

asif zaidi
Messages
56
Reaction score
0
Hello:

Problem1:
The temp of the circular plate D= {(x1,x2) | x1[tex]^{2}[/tex] + x2[tex]^{2}[/tex] [tex]\leq[/tex] 1} is given by T=2x[tex]^{2}[/tex] -3y[tex]^{2}[/tex] - 2x. Find hottest and coldest points of the plate.


Problem 2

Show that for all (x1,x2,x3) [tex]\in[/tex] R[tex]^{3}[/tex] with x1>0, x2>0, x3>0 and x1x2x3 = 1, we have x1+x2+x3 [tex]\geq[/tex]3


Solution to problem1
First I think there is a typo in the problem. D is given in terms of x1,x2 and T in terms of x,y. Shouldn't they both be in terms of either x1x2 or xy. If so then I have xy value s which I need to plug into T to find the max and min.

Solution to problem2:
This is where I am having real problems. I am not sure what my constraining function is.
What I have done so far is the following (and this is the crucial step which I may have gotten wrong). My question for this problem is at end.

Maximize a1+a2+a3 subject to a1a2a3 = 1. Formulating this gives me the following

i + j + k = [tex]\lambda[/tex](a2a3) i + [tex]\lambda[/tex](a1a3) j + [tex]\lambda[/tex](a1a2) k

Therefore

1 = [tex]\lambda[/tex](a2a3);
1 = [tex]\lambda[/tex](a1a3);
1 = [tex]\lambda[/tex](a1a2);

Multiplying lhs and rhs in above 3 gives me following

a1 = [tex]\lambda[/tex](a2a3)a1;
a2 = [tex]\lambda[/tex](a1a3)a2;
a3 = [tex]\lambda[/tex](a1a2)a3;

This gives me a1 = a2 = a3

Putting this above in constraint gives me

a1a1a1 = 1;
Therefore a1 = 1 = a2 = a3

So I am able to a1+a2+a3 = 3.
But I have not been able to prove a1+a2+a3>3. Any pointers.


Thanks

Asif
 
Physics news on Phys.org
Prob 1) Yep, typo.
Prob 2) If a1+a2+a3=3 is a minimum value of a1+a2+a3, the certianly a1+a2+a3>=3.
 
asif zaidi said:
Hello:

Problem1:
The temp of the circular plate D= {(x1,x2) | x1[tex]^{2}[/tex] + x2[tex]^{2}[/tex] [tex]\leq[/tex] 1} is given by T=2x[tex]^{2}[/tex] -3y[tex]^{2}[/tex] - 2x. Find hottest and coldest points of the plate.


Problem 2

Show that for all (x1,x2,x3) [tex]\in[/tex] R[tex]^{3}[/tex] with x1>0, x2>0, x3>0 and x1x2x3 = 1, we have x1+x2+x3 [tex]\geq[/tex]3


Solution to problem1
First I think there is a typo in the problem. D is given in terms of x1,x2 and T in terms of x,y. Shouldn't they both be in terms of either x1x2 or xy. If so then I have xy value s which I need to plug into T to find the max and min.
It could be a typo or they could just using different symbols to distinguish the function from the formula defining the region.

Solution to problem2:
This is where I am having real problems. I am not sure what my constraining function is.
What I have done so far is the following (and this is the crucial step which I may have gotten wrong). My question for this problem is at end.

Maximize a1+a2+a3 subject to a1a2a3 = 1. Formulating this gives me the following

i + j + k = [tex]\lambda[/tex](a2a3) i + [tex]\lambda[/tex](a1a3) j + [tex]\lambda[/tex](a1a2) k

Therefore

1 = [tex]\lambda[/tex](a2a3);
1 = [tex]\lambda[/tex](a1a3);
1 = [tex]\lambda[/tex](a1a2);

Multiplying lhs and rhs in above 3 gives me following

a1 = [tex]\lambda[/tex](a2a3)a1;
a2 = [tex]\lambda[/tex](a1a3)a2;
a3 = [tex]\lambda[/tex](a1a2)a3;

This gives me a1 = a2 = a3

Putting this above in constraint gives me

a1a1a1 = 1;
Therefore a1 = 1 = a2 = a3

So I am able to a1+a2+a3 = 3.
But I have not been able to prove a1+a2+a3>3. Any pointers.


Thanks

Asif[/QUOTE]

There are no "contraining functions". Neither of these is a "Laplace multiplier problem". You are asked to find the max and min of a given function on a given set. Those must occur:
1. In the interior of the set where the gradient is 0.
2. In the interior of the set where the gradient does not exist.
3. On the boundary of the set.

For problem 1, you find the gradient of T=2x2 -3y2 - 2x and determine where, if anywhere, it is 0 inside the given circle. Then "restrict" the function to the xcircle and find where the max and min are on that circle. One way to do that is to use Laplace multipliers. Another is to write x and y in terms of a parameter, say the angle [itex]\theta[/itex] around the circle.

For problem 2, you find the gradient of x+ y+ z and see where it is 0 in the interior of the set. For the boundary, just set x= 0, y= 0, z= 0 and xyz= 1 so z= 1/xy in your function. Not that the boundary includes the "corners"- the points of intersection of those sets.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K