Question

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

Answer 1

# {: ("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: