f(x,y)=xy(1−6x−5y)=xy−6x2y−5xy2.
The first order partial derivatives of f(x,y) are:
fx(x,y)=y−12xy−5y2=y(1−12x−5y),
fy(x,y)=x−6x2−10xy=x(1−6x−10y).
To find the critical points, we solve fx(x,y)=fy(x,y)=0.
We have:
y(1−12x−5y)=0 and x(1−6x−10y)=0.
The first equation is satisfied if either y=0 or 1−12x−5y=0.
The second equation is satisfied if either x=0 or 1−6x−10y=0.
Hence there are four possibilities:
1) y=0,x=0 and we have the critical point (0,0) .
2) y=0,1−6x−10y=0. In this case: 1−6x−12⋅0=0,1−6x=0,6x=1,x=61. We have the critical point (61,0).
3) x=0,1−12x−5y=0. In this case: 1−12⋅0−5y=0,1−5y=0,5y=1,y=51. We have the critical point (0,51).
4) 1−12x−5y=0,1−6x−10y=0.
12x+5y=1,6x+10y=1.
We multiply the first equation by 2 and subtract the second equation: (24x+10y)−(6x+10y)=2−1,18x=1,x=181.
Then we have from the first equation:
12⋅181+5y=1,32+5y=1,5y=31,y=151.
We have the critical point (181,151).
All four points in increasing lexicographic order: (0,0),(0,51),(181,151),(61,0).
To find the nature of the critical points, we can apply the second derivative test.
fxx(x,y)=−12y,fyy(x,y)=−10x,fxy(x,y)=1−12x−10y.
D=fxx(x,y)fyy(x,y)−(fxy(x,y))2=120xy−(1−12x−10y)2.
At the point (0,0):D=0−(1−0−0)2=0−12=−1<0, hence (0,0) is a saddle point.
At the point (0,51):D=0−(1−10⋅51)2=0−(1−2)2=−1<0, hence (0,51) is a saddle point of f.
At the point (181,151):D=120⋅181⋅151−(1−12⋅181−10⋅151)2=
=94−(1−32−32)2=94−(−31)2=94−91=93=31>0.
As fxx(181,151)=−12⋅151=−54<0, hence there is a local maximum at (181,151).
At the point (61,0):D=0−(1−12⋅61)2=0−(1−2)2=−1<0, hence (61,0) is a saddle point of f.
Answer.
First point (0,0) of type saddle.
Second point (0,51) of type saddle.
Third point (181,151) of type local maximum.
Fourth point (61,0) of type saddle.
