# Lagrange Multipliers - 2 questions

1. May 23, 2008

### asif zaidi

Hello:

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

Problem 2

Show that for all (x1,x2,x3) $$\in$$ R$$^{3}$$ with x1>0, x2>0, x3>0 and x1x2x3 = 1, we have x1+x2+x3 $$\geq$$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 = $$\lambda$$(a2a3) i + $$\lambda$$(a1a3) j + $$\lambda$$(a1a2) k

Therefore

1 = $$\lambda$$(a2a3);
1 = $$\lambda$$(a1a3);
1 = $$\lambda$$(a1a2);

Multiplying lhs and rhs in above 3 gives me following

a1 = $$\lambda$$(a2a3)a1;
a2 = $$\lambda$$(a1a3)a2;
a3 = $$\lambda$$(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

2. May 23, 2008

### benorin

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.

3. May 24, 2008

### HallsofIvy

Staff Emeritus
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 = $$\lambda$$(a2a3) i + $$\lambda$$(a1a3) j + $$\lambda$$(a1a2) k

Therefore

1 = $$\lambda$$(a2a3);
1 = $$\lambda$$(a1a3);
1 = $$\lambda$$(a1a2);

Multiplying lhs and rhs in above 3 gives me following

a1 = $$\lambda$$(a2a3)a1;
a2 = $$\lambda$$(a1a3)a2;
a3 = $$\lambda$$(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 $\theta$ 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.