How do you find the critical points for #f(x,y)=xy(1-8x-7y)#?

1 Answer
Jun 29, 2018

Answer:

There is a saddle point at #(0,0)# and a local maximum at #(1/24,1/21)#

Explanation:

The function is

#f(x,y)=xy(1-8x-7y)=xy-8x^2y-7xy^2#

Caculate the partial derivatives

#(delf)/(delx)=y-16xy-7y^2#

#(delf)/(dely)=x-8x^2-14xy#

The critical points are

#{(y-16xy-7y^2=0),(x-8x^2-14xy=0):}#

#<=>#, #{(y(1-16x-7y)=0),(x(1-8x-14y)=0):}#

Therefore, #(0,0)# is a point

#<=>#, #{((16x+7y)=1),((8x+14y)=1):}#

#<=>#, #{((16x+7y)=1),((16x+28y)=2):}#

#<=>#, #{(16x+7y=1),(y=1/21):}#

#<=>#, #{(x=1/24),(y=1/21):}#

The other point is #(1/24, 1/21)#

Calculate the second derivatives

#(del^2f)/(delx^2)=-16y#

#(del^2f)/(dely^2)=-14x#

#(del^2f)/(delxdely)=1-16x-14y#

#(del^2f)/(delydelx)=1-16x-14y#

Calculate the Determinant #D(x,y)# of the hessian Matrix

#((-16y,1-16x-14y ),(1-16x-14y,-14y))#

#D(x,y)=224y^2-(1-16x-14y)^2#

Therefore,

#D(0,0)=-1#

As #D(0,0)<0#, this is a saddle point.

#D(1/24,1/21)=0.51-0.11=0.4#

#D(1/24,1/21)>0#, then #(del^2f(1/24,1/21))/(delx^2)=-16/21#

#(del^2f(1/24,1/21))/(delx^2)<0#

This is a local maximum at #(1/24,1/21)#