Question

What are the local extrema an saddle points of `f(x,y) = x^2 + xy + y^2 + 3x -3y + 4`?

Answer 1

Please see the explanation below

Explanation:

The function is

`f(x,y)=x^2+xy+y^2+3x-3y+4`

The partial derivatives are

`(delf)/(delx)=2x+y+3`

`(delf)/(dely)=2y+x-3`

Let `(delf)/(delx)=0` and `(delf)/(dely)=0`

Then,

`{(2x+y+3=0),(2y+x-3=0):}`

`=>`, `{(x=-3),(y=3):}`

`(del^2f)/(delx^2)=2`

`(del^2f)/(dely^2)=2`

`(del^2f)/(delxdely)=1`

`(del^2f)/(delydelx)=1`

The Hessian matrix is

`Hf(x,y)=(((del^2f)/(delx^2),(del^2f)/(delxdely)),((del^2f)/(delydelx),(del^2f)/(dely^2)))`

The determinant is

`D(x,y)=det(H(x,y))=|(2,1),(1,2)|`

`=4-1=3 >0`

Therefore,

There are no saddle points.

`D(1,1)>0` and `(del^2f)/(delx^2)>0`, there is a local minimum at `(-3,3)`

Answer 2

Local minimum: `(-3,3)`

Explanation:

The group of points that include both extrema and saddle points are found when both `(delf)/(delx)(x,y)` and `(delf)/(dely)(x,y)` are equal to zero.

Assuming `x` and `y` are independent variables:
`(delf)/(delx)(x,y)=2x+y+3`
`(delf)/(dely)(x,y)=x+2y-3`

So we have two simultaneous equations, which happily happen to be linear:
`2x+y+3=0`
`x+2y-3=0`

From the first:
`y=-2x-3`
Substitute into the second:
`x+2(-2x-3)-3=0`
`x-4x-6-3=0`
`-3x-9=0`
`x=-3`
Substitute back into the first:
`2(-3)+y+3=0`
`-6+y+3=0`
`-3+y=0`
`y=3`

So there is one point where the first derivatives uniformly become zero, either an extremum or a saddle, at `(x,y)=(-3,3)`.

To deduce which, we must compute the matrix of second derivatives, the Hessian matrix (https://en.wikipedia.org/wiki/Hessian_matrix):
`(((del^2f)/(delx^2),(del^2f)/(delxdely)),((del^2f)/(delydelx),(del^2f)/(dely^2)))`

`(del^2f)/(delx^2)=2`
`(del^2f)/(delxdely)=1`
`(del^2f)/(delydelx)=1`
`(del^2f)/(dely^2)=2`

Thus
`(((del^2f)/(delx^2),(del^2f)/(delxdely)),((del^2f)/(delydelx),(del^2f)/(dely^2)))=((2,1),(1,2))`
All second order derivatives are uniformly constant whatever the values of `x` and `y`, so we do not need to specifically compute the values for the point of interest.

NB The order of differentiation does not matter for functions with continuous second derivatives (Clairault's Theorem, application here: https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives), and so we expect that `(del^2f)/(delxdely)=(del^2f)/(delydelx)`, as we see in our specific result above.

In this two-variable case, we can deduce the type of point from the determinant of the Hessian, `(del^2f)/(delx^2)(del^2f)/(dely^2)-(del^2f)/(delxdely)(del^2f)/(delydelx)=4-1=3`.

A form of the test to administer is given here:
https://en.wikipedia.org/wiki/Second_partial_derivative_test#The_test

We see that the determinant is `>0`, and so is `(del^2f)/(delx^2)`. So we conclude that `(-3,3)`, the sole point of zero first derivative, is a local minimum of the function.

As a sanity check for a one-dimensional function question, I usually post the graph of it, but Socratic does not have a surface or contour plotting facility suitable for two-dimensional functions, so far as I can see. So I will overplot the two functions `f(-3,y)` and `f(x,3)`, which do not characterise the whole function domain for us, but will show us the minimum between them, which appears as expected at `y=3` and `x=-3`, taking identical function value `f=-5` in each case.

As `f(x,y)=x^2+xy+y^2+3x-3y+4`
`f(-3,y)=y^2-6y+4`
`f(x,3)=x^2+6x+4`
graph{(x-(y^2-6y+4))(y-(x^2+6x+4))=0 [-10, 5, -6, 7]}