If f(x,y)=2y^2x-yx^2+4xy , find the local extrema and saddle points of f .

higher dimention problems are solved analogously with 2-D
extrema problems: take derivatives for each direction, and
solve.

Steps:

Find critical points
where Fx = Fy = 0; evaluate second derivative at those points:

. Fxx
< 0  & Fxx Fyy -  Fxy^2 > 0 -->
maximum

. Fxx < 0 & Fxx Fyy - Fxy^2 > 0
 --> minimum

. Fxx Fyy - Fxy^2 < 0  --> saddle
point

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

fx = 2y^2 - 2yx + 4y    -->  fxx = -2y  and  fxy = 4y -
2x + 4

fy = 4yx - x^2 + 4x   --> fyy = 4y   and  fyx = 4y -
2x + 4

Note that fxy = fyx, as it
should.

Solve system for critical
points:

fx = 2y^2 - 2yx + 4y = 0

fy
= 4yx - x^2 + 4x = 0

solutions: (0,-2), (4/3, -2/3), (4,0),
(0,0)

We'll complete the solution for one of these
points. Each of the points is evaluated in like manner:

fxx (4/3,
-2/3) = -2y = -2(-2/3) = 4/3 > 0

fyy(4/3, -2/3) = 4y =
4(-2/3) = -8/3

fxy(4/3, -2/3) = 4y - 2x + 4 = -8/3 - 8/3 + 4 =
-4/3

fxx fyy - fxy^2 = 4/3 * -8/3 - 16/9 = -16/3 <
0

Because fxx fyy - fxy^2 < 0, the point (4/3, -2/3) is a
saddle point.