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

*print*Print*list*Cite

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.