Assignment 13

35

f  =  -((((x^2) - 1))^2) -  ((((x^2) y  -  x  -  1))^2);

Critical Points:

cps = 
      Solve[{D[f, x]  ==  0,  D[f, y]  ==  0}]

d[x_, y_]  =  
      D[f, x, x] D[f, y, y] - D[f[x, y], x, y]^2 // Simplify

d[x, y]  /.  cps

D[f, x, x]  /.  cps

Since d  >  0 and (Subscript[f, xx]) < 0 at the critical points (-1, 0) and (1, 2), the function has local maxima at these points. There are no other critical points, so the graph has no saddle points or local minima.

Plot3D[f, {x, -2, 2}, {y, -1, 3}, PlotPoints -> 30,  
      BoxRatios -> Automatic, PlotRange -> {-5, 0}]

36

f =  3 x (E^y) -  (x^3) -  (E^(3  y));

Critical Points:

Solve[{D[f, x]  ==  0,  D[f, y]  ==  0}]

The only critical point with real coordinates is (1, 0).

d[x_, y_]  =  
      D[f, x, x] D[f, y, y] - D[f, x, y]^2 // Simplify

d[1, 0]

D[f, x, x]  /.  {x -> 1,  y -> 0}

Since d  >  0 and (Subscript[f, xx]) < 0 at the critical point (1, 0), the function has local maximum at this point. There are no other critical points, yet f does not have an absolute maximum at this point. For example, f(2, 0)  =  1 while f(-3, 0)  =  17.

Plot3D[f, {x, -2, 2}, {y, -4, 1}, BoxRatios -> Automatic, 
      PlotRange -> {-5, 5}]

The graph looks like it may have a saddle point, but the calculations above show there are no critical points other than the local maximum at (1, 0) so there cannot be a saddle point on the graph.