Question

What are the values and types of the critical points, if any, of `f(x,y) =x^-3 + e^x+y^2 - y^3`?

Answer 1

# {: ("Critical Point","Type"), ((-6.199,0,-0.002),"maximum"), ((-2.395,0,0.018),"saddle"), ((1.020,0,3.716),"maximum"), ((-6.199,2/3,0.146),"saddle"), ((-2.395,2/3,0.167),"minimum"), ((1.020,2/3,3.864),"saddle") :} #

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

So we have:

`f(x,y) = x^(-3) + e^x + y^2 - y^3`

Let us find the first partial derivatives:

`(partial f) / (partial x) = -3x^(-4)+e^x`

`(partial f) / (partial y) = 2y-3y^2`

So our critical equations are:

`e^x - 3/x^4 = 0` ..... [1]
`2y-3y^2 = 0` ..... [2]

From the these equations we have:

For [1] Using Newton Rhapson (to 4dp):
`x=-6.1988, -2.3951, 1.0199`

For [2]: `2y-3y^2 = 0 => y(2-3y)=0`
`y=0,2/3`

Any permutation of these solutions will simultaneously make both partial derivatives vanish, so the critical values are:

`(x,y) = (-6.199, 0), (-2.395, 0), (1.020, 0,)`
`" " = (-6.199, 2/3), (-2.395, 2/3), (1.020, 2/3)`

So we can now calculate the coordinates of the critical points:

# {: (f(-6.199,0),= -0.002,=>(-6.199,0,-0.002)), (f(-2.395,0),= 0.018,=>(-2.395,0,0.018)), (f(1.020,0),= 3.716,=>(1.020,0,3.716)), (f(-6.199,2/3),= 0.146,=>(-6.199,2/3,0.146)), (f(-2.395,2/3),= 0.167,=>(-2.395,2/3,0.167)), (f(1.020,2/3), = 3.864,=>(1.020,2/3,3.864)) :} #

We can visualise these critical points on a 3D-plot:

So, now let us look at the second partial derivatives so that we can determine the nature of the critical point (I'll just quote these results):

`\ \ \ (partial^2f) / (partial x^2) = e^x-12x^(-5)`

`\ \ \ (partial^2f) / (partial y^2) = 2-6y`

`(partial^2f) / (partial x partial y) = 0 \ \ \ \ (=(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"), ((-6.199,0,-0.002),0.001,2,0,>0,"maximum"), ((-2.395,0,0.018),-0.061,2,0,<0,"saddle"), ((1.020,0,3.716),13.647,2,0,>0,"maximum"), ((-6.199,2/3,0.146),0.001,-2,0,<0,"saddle"), ((-2.395,2/3,0.167),-0.061,-2,0,>0,"minimum"), ((1.020,2/3,3.864),13.647,-2,0,<0,"saddle") :} #