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Lagrange Multipliers to find max/min values

  • Thread starter arl146
  • Start date
  • #1
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1

Homework Statement


Use Lagrange multipliers to find the max and min values of the function subject to the given constraints:

f(x,y,z)= x2y2z2
constraint: x2 + y2 + z2 = 1


Homework Equations


∇f = ∇g * λ
fx = gx * λ
fy = gy * λ
fz = gz * λ

The Attempt at a Solution


i cant solve for x, y, and lambda

i got:
for this one, fx = gx * λ ---> 2x*y2*z2=(2x)λ
for, fy = gy * λ ---> 2y*x2*z2=(2y)λ
fz = gz * λ ----> 2z*x2*y2=(2z)λ

i tried setting them all equal to lambda, and ended up getting that x=y=z=1 but when you add that into g(x,y,z).... that equals >1 ? im confused with getting the x,y,z, and lambda values. help?
 

Answers and Replies

  • #2
343
1
oh please help!
 
  • #3
Ray Vickson
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Homework Statement


Use Lagrange multipliers to find the max and min values of the function subject to the given constraints:

f(x,y,z)= x2y2z2
constraint: x2 + y2 + z2 = 1


Homework Equations


∇f = ∇g * λ
fx = gx * λ
fy = gy * λ
fz = gz * λ

The Attempt at a Solution


i cant solve for x, y, and lambda

i got:
for this one, fx = gx * λ ---> 2x*y2*z2=(2x)λ
for, fy = gy * λ ---> 2y*x2*z2=(2y)λ
fz = gz * λ ----> 2z*x2*y2=(2z)λ

i tried setting them all equal to lambda, and ended up getting that x=y=z=1 but when you add that into g(x,y,z).... that equals >1 ? im confused with getting the x,y,z, and lambda values. help?

Assuming that x, y, z > 0 (other cases later) the first equation 2x*y^2*z^2= (2x)*u [using u instead of lambda] can be divided through by 2x on both sides to get y^2*z^2=u, that is: (yz)^2 = u. Similarly, (xz)^2 = u and (xy)^2 = u, so xz = xy = yz = v, v = sqrt(u) (since we have assumed all x, y, z > 0). So we get x = y = z. Now use the constraint.

Note that there are lots of other stationary points, essentially differing by signs in each of x, y and z. Also: you need to check whether you can have solutions with, for example, x = 0.


RGV
 
  • #4
I think I'm working on something similar.... this is what I would have done

[tex]\begin{gathered}
f\left( {x,y,z} \right) = {x^2}{y^2}{z^2};g\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 1 = 0 \\
\vec \nabla f = \left\langle {2x{y^2}{z^2},2{x^2}y{z^2},2{x^2}{y^2}z} \right\rangle ;\vec \nabla g = \left\langle {2x,2y,2z} \right\rangle \\
\vec \nabla f = \lambda \vec \nabla g \to {y^2}{z^2} = {x^2}{z^2} = {x^2}{y^2} \to x = \pm y = \pm z \\
3{x^2} = 1 \to {x^2} = \frac{1}{3} \to x = \pm \frac{{\sqrt 3 }}{3} \\
f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( {\frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = \\
f\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3} - \frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3},\frac{{\sqrt 3 }}{3}} \right) = f\left( { - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}, - \frac{{\sqrt 3 }}{3}} \right) = \frac{1}{{27}} \\
\end{gathered}[/tex]
 
  • #5
343
1
Assuming that x, y, z > 0 (other cases later) the first equation 2x*y^2*z^2= (2x)*u [using u instead of lambda] can be divided through by 2x on both sides to get y^2*z^2=u, that is: (yz)^2 = u. Similarly, (xz)^2 = u and (xy)^2 = u, so xz = xy = yz = v, v = sqrt(u) (since we have assumed all x, y, z > 0). So we get x = y = z. Now use the constraint.

Note that there are lots of other stationary points, essentially differing by signs in each of x, y and z. Also: you need to check whether you can have solutions with, for example, x = 0.


RGV
i did that whole thing... i got that x=y=z and so im confused because how does x^2 + y^2 + z^2 = 1 if x=y=z ?
 
  • #6
343
1
arcyqwerty: didnt even think about doing that way the 3x^2=1
 
  • #7
343
1
so is arcyqwerty right? haha is there really those 8 points and only those?
 
  • #9
343
1
Last edited by a moderator:
  • #10
343
1
going off this, but using it for another problem .. i think i see a trend...
whatever 'n' is, is what the values are. like how we found that x=y=z= 1/sqrt(3) so not using values, just using 'n', we can assume that each value will always equal 1/sqrt(n) with just these types of problems.... where you have f(x1, x2, ... , xn)= x1+x2+...+xn and the constraint is x1^2 + x2^2 + ... + xn^2
 

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