# Lagrange multiplier systems of equations -- Help please

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1. Oct 28, 2014

### tix24

1. The problem statement, all variables and given/known data
Hi guys im new here and i really need help with this question. Ive tried it multiple times but cant find all the critical points, help would be greatly appreciated.

the question is as follows:
Find the maximum and minimum values of w=4x-(1/2)y+(27/2)z on the surface (x^4)+(y^4)+(z^4)=1.
2. Relevant equations

since we have a constraint we have to use the lagrange multiplier method.

3. The attempt at a solution

i dont really know how to write all the stuff here;( note: lambda is denoted by L, i have no idea how to put symbols on this post) but i got the system 4=4λ(x^3), -0.5=4λ(y^3), (27/8)=λ(z^3) and the constraint equations which is also in the system (X^4)+(y^4)+(z^4)=1.

i got the points
x=(16/98)^1/4
x2=-(16/98)^1/4
y=(1/98)^1/4
y2=-(1/98)^1/4
z=(81/98)^1/4
z2=(81/98)^1/4

2. Oct 28, 2014

### Dick

It looks good so far except you forgot a '-' on z2. So that means you have eight critical points (x,y,z), yes? Two possibilities for each coordinate number.

3. Oct 28, 2014

### Ray Vickson

Often, in the Optimization world, people use symbols like u, v, etc. for Lagrange multipliers, instead of $\lambda$, partly because it is easier to type. So, that is what I will do.

In problems like this you also need to distinguish between maxima and minima. It is easy enough in this case, just by evaluating w at the two points.

It is obvious (and easy to prove in this case) that the solution is the same as the max or min problem on the full 3-dimensional constraint set x^4 + y^4 + z^4 <= 1 (because the solution of the latter must occur on the boundary---just by the nature of w(x,y,z)). The advantage of recognizing this is that we can know automatically the sign of the Lagrange multiplier, u. Setting g = x^4+y^4+z^4-1, the constraint is g <= 0.

In the maximization we can set the Lagrangian to L_max = w -u*g; then we know from theory that we need u >= 0. In the minimization problem we can set the Lagrangian to L_min = w + u*g; then we know we must have u >= 0 also. Alternatively, you can just use a single Lagrangian L = w - u*g and note that u >= 0 in the max problem, but u <= 0 in the min problem. (These facts depend on some material you might not have seen yet, connected with the so-called Karush-Kuhn-Tucker conditions.)

Anyway, some of your signs are wrong in your solutions above (but the magnitudes are correct). I suggest you re-solve the equations, paying very close attention to signs. Also, make sure you identify which x's go with which y's and which z's, etc. If you show the details of your work we can be more helpful.

4. Oct 28, 2014

### tix24

yea sorry, i got it to be a negative value but didnt put it on the thread (accident)
i have no idea what to do now though, ive been trying to do this problem for two days and i get stuck at this point, can you please guide me through?

5. Oct 28, 2014

### tix24

how do i know which xs go with which ys and zs? im really stuck on this problem. With regard to what you wrote above, my textbook just talks about using the lagrange multiplier method, it says to find the candidates/critical points using the lagrange method then to plug the values into f so that we get values for each point. then the largest is our max and our smallest is our min. I dont know which values of x,y,z go with each other. can you help me out?

6. Oct 28, 2014

### Ray Vickson

You must show your work; otherwise I cannot be helpful. I need to see what you are doing so I can comment on various steps (if it is appropriate for me to do so).

7. Oct 28, 2014

### Dick

I was being sloppy. There aren't that many critical points. There are relations between the signs of the coordinates. As Ray suggests show your work and think it over.

8. Oct 28, 2014

### tix24

so i just resolved the problem again and i got 8 critical points, i got (x1,y1,z1) (x1,y1,z2) (x1,y2,z1) (x1,y2,z2) (x2,y2,z2) (x2,y2,z1) (x2,y1,z1) (x2,y1,z2)

9. Oct 28, 2014

### tix24

can any body guide me or confirm this?

10. Oct 28, 2014

### tix24

Here is the work

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11. Oct 28, 2014

### tix24

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12. Oct 28, 2014

### Ray Vickson

There are just two critical points. I cannot figure out what you did wrong. I will not look at posted pictures of work instead of typed work.

13. Oct 28, 2014

### tix24

so when i found all of my partial derivatives and equated everything i obtained the following system

λ=1/(x)^3
λ=-1/8(y)^3
λ=27/8(z^3)
x^4+y^4+z^4=1

i then solved for y in terms of x and obtained y=(-x^3/8)^(1/3)
next i solved for z in terms of x and got z=(27x^3/8)^(1/3)

plugging these values into x^4+y^4+z^4=1 gave me the result

x=(8/49)^(1/4) and x=-((8/49)^(1/4))

i know have 2 values for x

i went back to my 4 equations
λ=1/(x)^3
λ=-1/8(y)^3
λ=27/8(z^3)
x^4+y^4+z^4=1

and now solved x and z in terms of y

x=-2y and z=3y

i plugged these values in x^4+y^4+z^4=1 and solved for y

i got 2 y values which are y=(1/98)^(1/4) and y=-((1/98)^(1/4))

now that i have values for x and y i just need to find z

so i i do the same procedure as i did for x and y and obtain two values for z

z=(81/98)^(1/4) and z=-((81/98)^(1/4))

i now have 2 values for x, 2 for y and 2 for z

x1=(8/49)^(1/4) x2=-((8/49)^(1/4))
y1=(1/98)^(1/4) y2=-((1/98)^(1/4))
z1=(81/98)^(1/4) z2=-((81/98)^(1/4))

the possible combinations of the points above are

x1,y1,z1) (x1,y1,z2) (x1,y2,z1) (x1,y2,z2) (x2,y2,z2) (x2,y2,z1) (x2,y1,z1) (x2,y1,z2)

so these are the points that i found.

14. Oct 28, 2014

### Ray Vickson

Well, you shouldn't have! Note: my " technical" response appears in blue above; I don't know why it is in a different panel.

Last edited: Oct 28, 2014