What are the extrema and saddle points of #f(x,y) = 2x^3 + xy^2 + 5x^2 + y^2#?

1 Answer
Feb 17, 2017

Answer:

# {: ("Critical Point","Conclusion"), ((0,0), "min"), ((-1,-2), "saddle"), ((-1,2), "saddle"), ((-5/3,0),"max") :} #

Explanation:

The theory to identify the extrema of #z=f(x,y)# is:

  1. Solve simultaneously the critical equations

    # (partial f) / (partial x) =(partial f) / (partial y) =0 \ \ \ # (ie #z_x=z_y=0#)

  2. Evaluate #f_(x x), f_(yy) and f_(xy) (=f_(yx))# at each of these critical points. Hence evaluate #Delta=f_(x x)f_(yy)-f_(xy)^2# at each of these points
  3. Determine the nature of the extrema;

    #{: (Delta>0, "There is minimum if " f_(x x)<0),(, "and a maximum if " f_(yy)>0), (Delta<0, "there is a saddle point"), (Delta=0, "Further analysis is necessary") :}#

So we have:

# f(x,y) = 2x^3+xy^2+5x^2+y^2 #

Let us find the first partial derivatives:

# (partial f) / (partial x) = 6x^2+y^2+10x#
# (partial f) / (partial y) = 2xy+2y#

So our critical equations are:

# 6x^2+y^2+10x = 0#
# 2xy+2y = 0#

From the second equation we have:

# 2y(x+1) = 0 => x=-1,y=0#

Subs #x=-1# into the First equation and we get:

# 6+y^2-10 = 0 => y^2=4=>y=+-2#

Subs #y=0# into the First equation and we get:

# 6x^2+0^2+10x = 0 => 2x(3x+5) =0 => x=-5/3,0#

And so we have four critical points with coordinates;

# (-1,-2), (-1,2), (0,0), (-5/3,0) #

So, now let us look at the second partial derivatives so that we can determine the nature of the critical points:

# \ \ \ (partial^2f) / (partial x^2) = 12x+10#
# \ \ \ (partial^2f) / (partial y^2) = 2x+2#
# (partial^2f) / (partial x partial y) = 2y \ \ \ \ (=(partial^2f) / (partial y partial x))#

And we must calculate:

#Delta=(partial^2f) / (partial x^2) (partial^2f) / (partial y^2) - ((partial^2f) / (partial x partial y))^2 #

at each critical point. The second partial derivative values, #Delta#, and conclusion are as follows:

# {: ("Critical Point",(partial^2f) / (partial x^2),(partial^2f) / (partial y^2),(partial^2f) / (partial x partial y),Delta,"Conclusion"), ((0,0),10,2,0,gt 0,f_(x x)>0 => "min"), ((-1,-2),-2,0,4,lt 0,"saddle"), ((-1,2),-2,0,4,lt 0,"saddle"), ((-5/3,0),-10,-4/3,0,gt 0,f_(x x)<0 => "max") :} #

We can see these critical points if we look at a 3D plot:

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