**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)