What are the local extrema an saddle points of #f(x,y) = x^2 + xy + y^2 + 3x -3y + 4#?
2 Answers
Please see the explanation below
Explanation:
The function is
The partial derivatives are
Let
Then,
The Hessian matrix is
The determinant is
Therefore,
There are no saddle points.
Local minimum:
Explanation:
The group of points that include both extrema and saddle points are found when both
Assuming
So we have two simultaneous equations, which happily happen to be linear:
From the first:
Substitute into the second:
Substitute back into the first:
So there is one point where the first derivatives uniformly become zero, either an extremum or a saddle, at
To deduce which, we must compute the matrix of second derivatives, the Hessian matrix (https://en.wikipedia.org/wiki/Hessian_matrix):
Thus
All second order derivatives are uniformly constant whatever the values of
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
In this two-variable case, we can deduce the type of point from the determinant of the Hessian,
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
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
As
graph{(x-(y^2-6y+4))(y-(x^2+6x+4))=0 [-10, 5, -6, 7]}