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 Check out our web site! http://mathforum.org/dr.math/
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