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Finding the Maximum and Minimum Values

  • #1

Homework Statement



Find the maximum and minimum values of 3x+3y+2z on the sphere x^2+y^2+z^2=1 .

Homework Equations



Use of LaGrange Multipliers (maybe?)

The Attempt at a Solution



I have no idea if LaGrange multipliers is the way to go or not, but I took the partial derivative for x, y and z

Partial x = 3
Partial y = 3
Partial z = 2

Using LaGrange Multipliers
I then got the following equations:
3=lambda2x
3=lambda2y
2=lambda2z

Lambda being the LaGrange Multiplier I believe.

I also have the equation of the sphere or x^2+y^2+z^2=1

However, from here I had no idea where to go next. I tried to solve out the LaGrange Formulas but couldn't get a coherent answer. Am I on the right track or do I need to just forget LaGrange?
 

Answers and Replies

  • #2
1,013
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You are on the correct track. Where specifically are you running into a problem?
 
  • #3
I am having trouble solving out my lagrange multipliers. Like if I solve it out, I think I get lambda being 3/2 leaving x and y as 1 and z as 2/3.
 
  • #4
1,013
70
I am having trouble solving out my lagrange multipliers. Like if I solve it out, I think I get lambda being 3/2 leaving x and y as 1 and z as 2/3.
I don't really see how you get that. Solving for x, y, and z should give you x = 3L/2, y = 3L/2, z = L. Then you can use these values in the restraint (the equation of the sphere) to solve for L.
 
  • #5
I don't really see how you get that. Solving for x, y, and z should give you x = 3L/2, y = 3L/2, z = L. Then you can use these values in the restraint (the equation of the sphere) to solve for L.
I had always thought that you were supposed to just be able to look at the equations and find lambda or what not because that's what the teacher had done, but this way actually makes sense.

So I solved out the L in the equation of the sphere and got sqrt(4/22)

Plugging it back into my equations I solved for x, y and z and got:

x= 3sqrt(22)/4
y= 3sqrt(22)/4
z= sqrt(4/22)

from here I just plug in these values into the original function to find the maximum and minimum values right?
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
Well, substituting back into the constraint, I get L^2=4/22=2/11. Doesn't that have more than one solution? You should probably expect to get two, since there ought to be a max and a min. There seems to be some flipping of L and 1/L going on here, which is ok, but it would be nice to stick with one. Can you show your complete solution, CloudTenshi?
 
  • #7
I just realized I made a few typos on the last one, but after solving for the x, y, and z using the value I got for L which was the sqrt(11/2) and -sqrt(11/2). From this, I was able to get the x, y, and z values which are:

x=3/(2L) = plus or minus (3/2)(sqrt(2/11))
y=3/(2L) = plus or minus (3/2)(sqrt(2/11))
z=1/L = plus or minus (sqrt(2/11))

I then plugged the the values back into the original function to find the max, which is when everything is positive, and min which is when everything is negative.

I think I should have the right answer.

The max is 11sqrt(2/11)
and min is -11sqrt(2/11)
 
  • #8
Dick
Science Advisor
Homework Helper
26,258
618
Yeah, I think you got it.
 
  • #9
Thanks. I think it all became clear for me when I found out that I was supposed to solve for L and then for the x,y,z coordinates.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,833
956
By the way, it is often possible to solve max-min problems, using Lagrange multipliers, without ever finding the multiplier itself! For example, you have
[itex]3=\lambda 2x[/itex], [itex]3=\lambda 2y[/itex], and [itex]2=\lambda 2z[/itex].

Dividing the second equation by the first eliminates [itex]\lambda[/itex] giving 1= y/x and dividing the third equation by the first gives 2/3= z/x so that y= x and z= (2/3)x. Put those into the equation of the sphere to solve for x, then find y and z.
 

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