What are the maximum and minimum values of f(x,y,z) = xyz subject to the constraint x+y+z =1, and find the points at which these extreme values occur.

Question

Assume x,y,z => 0

 

Apologies, it cut off. The inequality is the part that trips me up

 

For minimum I answered that the min 0 is at x = 0 and y+z=1, y =0 and x+z=1, z=0 and x+y=1. However the correct location was (1,0,0), (0,1,0) and (0,0,1). I understand Lagrange multipliers but please help me clarify where my reasoning went awry.

Answer 1

The method of Lagrange multipliers can be used to find extrema lying in the interior of some domain. 

It appears that for this problem, the domain is the first octant where x,y,z are non negative.

 

A straightforward application of the Lagrange method finds the value 1/27 at (1/3 , 1/3 , 1/3).   A bit of exploring the neighborhood of this point shows that it is a maximum. 

 

The minimum is clearly zero because of the restriction to the first octant.     A large set of points on the three bounding surfaces of the first octant will produce xyz = 0.

 

So both your answer and the "correct" answer are right.   However both are incomplete. 

Answer 2

by taking partial derivatives we get f_x = yz, f_y = xz, f_z = xy and g_x=g_y=g_z = 1 and so we solve the system

I.  yz = λ(1)

II. xz = λ(1)

III. xy = λ(1).

 

I'll assume that all of x, y and z must be non-negative; OK?

 

Thus (I, II both = λ(1) 0, yz = xz  so x=y.

(II,III equated) implies xz = xy    so z = y

(I,III equated) implies yz = xy     so x = z.

 

It's also possible that any two of the variables could be zero, in which case the third would have to be 1. This means that lambda would have to be zero, so those considerations are apart from the above steps I-III, where non-zero lambda is assumed. 

 

Because of the g(x,y,z) = x+y+z-1=0, we have x=y=z=1/3 each.  This should be the maximum.

 

Feedback will be appreciated.