2 questions regarding Lagrange Multipliers: 1) f(x,y,z)=x^2+y^2+z^2 and g(x,y,z)

Question

2 questions regarding Lagrange Multipliers:

1) f(x,y,z)=x^2+y^2+z^2 and g(x,y,z)=x+y+z-12

I found that x=y=z=4, so f(4,4,4)=48. However, I don't know if this is a max or a min, or how to find the other global value.

2) f(x,y,z)=x^2*y^2*z^2 and g(x,y,z)=x^2+y^2+z^2-1

Very similar to the last problem, I found that x^2=y^2=z^2=+-1/sqrt(3) but do not know how to find the other global value.

I have heard about just plugging in 0 to some of the equations but my teacher never mentioned this as an option. What do I do to find the second global value so I can determine what is a Global Minimum and Global Maximum?

Explanation / Answer

Solution:

1)

Given f(x,y,z)=x^2+y^2+z^2 and g(x,y,z)=x+y+z-12

f = g ==> <2x, 2y, 2z> = <1, 1, 1>.

So, = 2x = 2y = 2z by equating like entries.
==> x = y = z.

Substituting into g yields 3x = 12 ==> x = 4.

So, we have one critical point (4, 4, 4), and f(4, 4, 4) = 48.
This is minimal, because any other point satisfying g will yield a smaller value from f:
For instance, (0, 0, 12) satisfies g, but f(0, 0, 12) = 144 > 48

2)

Given f(x,y,z)=x^2*y^2*z^2 and g(x,y,z)=x^2+y^2+z^2-1

f = g ==> <2xy^2 z^2, 2x^2 y z^2, 2x^2 y^2 z> = <2x, 2y, 2z>.

So, 2x^2 y^2 z^2 = 2x^2 = 2y^2 = 2z^2 by equating like entries.

If = 0, then at least one of x, y, z = 0, and f(x,y,z) = 0, which is the minimum
(since f is a product of squares).

Otherwise, we have x^2 = y^2 = z^2.
Substituting into g yields 3x^2 = 1 ==> x^2 = 1/3.

Hence, we have 8 critical values (accounting for all possible signage combinations):
(x, y, z) = (±1/3, ±1/3, ±1/3), and all of these yield the maximum value 1/27.

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