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

2 Answers
Jun 17, 2018

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)

Jun 17, 2018

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]}