Min, Max of 3-Variable Function

Date: 8/16/96 at 13:7:16
From: Alan
Subject: Min, Max of 3-Variable Function

Find all the critical points and determine their nature for the 
function z = x^3 - 6xy + y^3.

Date: 8/17/96 at 16:2:30
From: Doctor Anthony
Subject: Re: Min, Max of 3-Variable Function

For max, min and saddle points of the function f(x,y) we require:
partial_df/dx = 0   and partial_df/dy = 0

Now using the notation fxx to mean partial_d^2f/(dx)^2 etc, the 
following conditions will determine whether the point is a max, 
a min, or a saddle point.

Let D = (fxy)^2 - fxx*fyy then if: 

D<0 and fxx<0 we have a maximum
D<0 and fxx>0 we have a minimum
D>0 we have a saddle point.

If D = 0 then more refined tests are required to determine the nature 
of the point.

Consider now the question that has been set.

f(x,y) = x^3 - 6xy + y^3

fx = 3x^2 - 6y        fy = 3y^2 - 6x

Put each of these equal to zero to find the stationary points.

  x^2 - 2y = 0 ...(1)     and y^2 - 2x = 0 ...(2)   

From (2) 2x = y^2, so x = (y^2)/2  put this in (1) and we get

     (y^4)/4 - 2y = 0     so y = 0 (and therefore also x = 0) or

          (y^3)/4 = 2

              y^3 = 8   and y = 2   x = 4/2 = 2

Stationary points are at (0,0)  and (2,2)

To determine the nature we now proceed to find fxx, fxy and fyy.

fxx = 6x     fxy = -6      fyy = 6y

D = (fxy)^2 - fxx*fyy = 36 - 36xy

At (0,0) D = 36 (>0)  So this is a saddle point.

At (2,2) D = 36 - 36*4 = 36(1 - 4) = -3*36 = - 108
                                   also fxx = 12

So we D<0 and fxx>0 and this is the condition for a minimum point.

-Doctor Anthony,  The Math Forum
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