Lagrange Multipliers

[Guest]

Use Lagrange multipliers to find the max and min values of the function subject to the given constraint(s).

f(x,y,z)= x^2*y^2*z^2 ; x^2 + y^2 + z^2 =1

*(lm) stands for the lagrange multiplier symbol.

I did:
fx= 2xy^2z^2 = (lm)2x
fy= 2x^2yz^2 = (lm)2y
fz= 2x^2y^2z = (lm)2z

(lm) = (2xy^2z^2)/(2x) = (2x^2yz^2)/(2y) = (2x^2y^2z)/(2z)
or
x^2=y^2=z^2
and
x^2+y^2+z^2 = 1

If that is correct so far, here is where i am stuck. I'm not sure if the possible values would be (+-1, +-1, +-1) or i sub x^2 into all values to get:
x^2 + x^2 + x^2 = 1
x= sqrt(1/3)
and the expected values would be (+-sqrt(1/3), +-sqrt(1/3), +-sqrt(1/3))

If the possible values were (+-1, +-1, +-1)
then i got:
max - f(1,1,1) = 1
min - f(-1,-1,-1) = -1

Am i even close here?

 

[soroban]

Hello, almostacasper!

I'd say your work is correct,
. . but I have a different conclusion.

Use Lagrange multipliers to find the max and min values of the function subject to the given constraint(s).

. . \(\displaystyle f(x,y,z)\:=\:x^2y^2z^2,\;\; x^2\,+\,y^2\,+\,z^2\:=\:1\)

Observation: The points are limited to the surface of a unit sphere.
We wish to maximize/minimize the product of the squares of the coordinates.
Note that \(\displaystyle f(x,y,z)\:\geq\:0\)


You did:
. . \(\displaystyle \begin{array}{cc}f_x\:=\: 2xy^2z^2\:=\:2\lambda x\\
f_y\:=\: 2x^2yz^2\:=\:2\lambda y \\
f_z\:=\:2x^2y^2z\:=\:2\lambda z \end{array}\)

. . Then: \(\displaystyle \:\lambda\:=\:\frac{2xy^2z^2}{2x}\:=\:\frac{2x^2yz^2}{2y}\:=\:\frac{2x^2y^2z}{2z}\;\) [1]

. . or: \(\displaystyle \:x^2\:=\:y^2\:=\:z^2\,\) and \(\displaystyle \,x^2\,+\,y^2\,+\,z^2\:=\:1\;\) Correct!


Your second suggestion is correct:

Substitute and get: \(\displaystyle \:x^2\,+\,x^2\,+\,x^2\:=\:1\;\;\Rightarrow\;\;x\:=\:\pm\frac{1}{\sqrt{3}}\)

. . The critical points are: \(\displaystyle \:\left(\pm\frac{1}{\sqrt{3}},\:\pm\frac{1}{\sqrt{3}},\:\pm\frac{1}{\sqrt{3}}\right)\)

Then: \(\displaystyle \:f\left(\pm\frac{1}{\sqrt{3}},\,\pm\frac{1}{\sqrt{3}},\,\pm\frac{1}{\sqrt{3}}\right) \:=\:\left(\pm\frac{1}{\sqrt{3}}\right)^2\left(\pm\frac{1}{\sqrt{3}}\right)^2\left(\pm\frac{1}{\sqrt{3}}\right)^2 \:=\:\frac{1}{27}\;\) maximum

. . for any of the eight permutations of the signs.


Are there any minimum points? . . . Yes!

In [1], we cancelled \(\displaystyle x\)'s, assuming that \(\displaystyle x\,\neq\,0.\)

Suppose \(\displaystyle x\,=\,0.\;\) Then \(\displaystyle y^2\,+\,z^2\:=\:1\)
. . We have a unit circle on the \(\displaystyle yz\)-plane.

Similarly, if \(\displaystyle y\,=\,0\), we have a unit circle on the \(\displaystyle xz\)-plane.
And if \(\displaystyle z\,=\,0\), we have a unit circle on the \(\displaystyle xy\)-plane.

For all those points, \(\displaystyle f(0,y,z) \:=\:f(x,0,z) \:=\:f(x,y,0) \:=\:0\;\) minimum