Question

What are the values and types of the critical points, if any, of `f(x, y) = xy^2+xy-3x^2y-7x`?

Answer 1

Saddle points at `(0,-1/2-sqrt(29)/2)` and `(0,-1/2+sqrt(29)/2)`

Explanation:

We have a 3D surface defined by:

`f(x, y) = xy^2 + xy - 3x^2y - 7x`

Step 1 - Find the Partial Derivatives

We compute the partial derivative of a function of two or more variables by differentiating wrt one variable, whilst the other variables are treated as constant. Thus:

The First Derivatives are:

`f_x \ = (partial f) / (partial x) \ \ = y^2+y-6xy -7`
`f_y \ = (partial f) / (partial y) \ \ = 2xy+x-3x^2`

The Second Derivatives are:

`f_(x x) =(partial^2 f) / (partial x^2) = -6y`
`f_(yy) = (partial^2 f) / (partial y^2) = 2x`

The Second Partial Cross-Derivatives are:

`f_(xy) =(partial^2 f) / (partial x partial y) =2y+1-6x`
`f_(yx) = (partial^2 f) / (partial y partial x) =2y+1-6x`

Note that the second partial cross derivatives are identical due to the continuity of `f(x,y)`.

Step 2 - Identify Critical Points

A critical point occurs at a simultaneous solution of

`f_x = f_y = 0 iff (partial f) / (partial x) = (partial f) / (partial y) = 0`

i.e, when:

`y^2+y-6xy -7 = 0` ..... [A]
`2xy+x-3x^2 = 0` ..... [B]

From [B], we have:

`x(2y+1-3x) = 0 => x=0,1/3(2y+1)`

From [A], we have:

`x=0 => y^2+y -7 = 0`
`:. y = -1/2+-sqrt(29)/2`

`x=1/3(2y+1) => y^2+y-6(1/3(2y+1))y -7 = 0`
`:. y^2+y-4y^2-2y -7 = 0`
`:. 3y^2+y +7 = 0` having no real solutions

So we can conclude that there are two critical points:

`(0,-1/2-sqrt(29)/2)` and `(0,-1/2+sqrt(29)/2)`

Step 3 - Classify the critical points

In order to classify the critical points we perform a test similar to that of one variable calculus using the second partial derivatives and the Hessian Matrix.

`Delta = H f(x,y) = | ( f_(x x) \ \ f_(xy) ) , (f_(yx) \ \ f_(yy)) | = | ((partial^2 f) / (partial x^2),(partial^2 f) / (partial x partial y)), ((partial^2 f) / (partial y partial x), (partial^2 f) / (partial y^2)) | = f_(x x)f_(yy)-(f_(xy))^2`

Then depending upon the value of `Delta`:

`{: (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") :}`

Using custom excel macros the function values along with the partial derivative values are computed as follows:

And we can confirm these results graphically