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

Answer:

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

Answer:

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]}