Find the local max/min values and saddle points of the function f(x,y)= xy(1-x-y)

local-maxmin-values
saddle-points

#1

find the local max/min values and saddle points of the function f(x,y)= xy(1-x-y)

Answer:

fx/dx = y − 2xy − y^2
= y(1 − 2x − y),
fy/dy = x − 2xy − x^2
= x(1 − 2y − x).
Making fx and fy both zero, there
could be four possible cases : (0, 0), (1, 0), (0, 1), ( 1/3,1/3)
. Now, let’s compute second partial derivatives.
fxx/dx = −2y,
fxy/dy = 1 − 2x − 2y,
fyy/dy = −2x.
Hence, D = fxxfyy − f^2xy < 0 for (0, 0), (1, 0), (0, 1). So those points are saddle points.
However, for
( 1/3,1/3), it is positive and fxx < 0. Thus a local maximum is attained at the point ( 1/3,1/3).
Answer:
f has its local maximum 1
27 at ( 1/3,1^3)
and the saddle points are (0, 0), (1, 0), (0, 1)