Linear Systems and Quadratic Extrema
Many applications involve quadratic functions, where a
quadratic function is a function that is a second degree polynomial in each
variable. When a quadratic function has a critical point, it must be the
solution to a system of simultaneous linear equations (also known as a
linear system) of the form
One way of solving a linear system is to multiply the first equation by
either c or d, multiply the second equation by either -a or -b, and
then combine the two equations to eliminate one of the variables:
| ad x+bd y |
= |
r d |
| -bcx-bdy |
= |
-s b |
 |
|
|
|
( ad-bc) x |
= |
rd-sb |
|
The result can be solved for x, and then y can be found by either
repeating the process or substituting for x in one of the equations.
EXAMPLE 4 Find the extrema and saddle points of
|
f( x,y) = x2+3xy-y2-2x-3y |
|
Solution: To begin with, the first partial derivatives are
|
fx( x,y) = 2x+3y-2, fy( x,y) = 3x-2y-3 |
|
and thus, the critical point(s) must satisfy the linear system
We multiply the first equation by 2, the second by 3, and combine the
results:
As a result, we have x = 1, which after substituting into 2x+3y = 2 shows us
that 2+3y = 2 or y = 0. Thus, the critical point of f( x,y) is
( 1,0) .
The second derivatives of f( x,y) are
|
fxx( x,y) = 2, fyy = -2, fxx = 3 |
|
so that the discriminant of f( x,y) is
|
D = 2( -2) -( -3) 2 = 4-9 = -5 < 0 |
|
As a result, f( x,y) has a saddle at ( 1,0) .
The second derivative test is very important in applications, and consequently,
it appears in a wide variety of settings.
EXAMPLE 5 Find the point(s) on the plane z = x+y-3 that are
closest to the origin.
Solution: To begin with, we let f denote the square of
the distance from a point ( x,y,z) to the origin.
Consequently,
Substituting z = x+y-3 thus yields
|
f( x,y) = x2+y2+( x+y-3) 2 |
|
Since fx = 4x+2y-6 and fy = 2x+4y-6, we must solve
Multiplying the second equation by -2 yields
so that y = 1. Similarly, we find that x = 1, so the critical point is ( 1,1) . Moreover, fxx = 4, fxy = 2, and fyy = 4, so
that the discriminant is
|
D = fxxfyy-fxy2 = 16-4 = 12 > 0 |
|
Thus, every "slice'' is concave up and correspondingly, f has a minimum
at ( 1,1) . Substitution yields
so that ( 1,1,-1) is the point in the plane z = x+y-3 that is
closest to the origin.
Check your reading: Why did we use the square of the distance
instead of the actual distance in example 5?