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Question:-
Let be a function f(x,y)=3x2 - x+ xy2 +2 of two variables
defined on R2 with continuous second order
partial derivatives. Determine whether or not f has a relative
minimum, a relative maximum and a saddle point in R2.
Solution:-
Solution:
f (x, y)
=3x2 - x+ xy2 +2
f (x)¢ =3(2x) -1+ y2 +0
f (x)¢ = 6x-1+ y2 --------A
f (y)¢ = 0-0+ x(2y)
+0
f (y)¢ = 2xy -------B
Critical point of f (x, y);
f (x)¢ =0
And
f (y)¢ =0
From A and B;
6x-1+ y2 =0-------C
2xy =0------D
From D;
x = 0
y = 0
put x =0in C;
6(0) -1+ y2 = 0
0-1+ y2 = 0
y2 -1= 0
y2 =1
By Taking Square Root on both sides;
y =±1
So, points are (0,1)and (0,-1)
Now Put y =0 in C;
6x-1+
y2 = 0
6x-1+(0)2
=0
6x-1=0
6x =1
x =1/ 6
Points are (1/ 6,0)
Taking again derivatives of equation A and B;
f (x)¢ =6x-1+ y2 --------A
f (x)¢¢ =6(1)-0+0
f (x)¢¢ =6
f (y)¢ = 2xy -------B f (y)¢¢ = 2x(1)
f (y)¢¢ = 2x
Now;
f (x)¢ =6x-1+ y2 --------A f (x)¢y =0-0+2y
f (x)¢y = 2y
Also
Now putting values of critical points (0,1)and (0,-1)
f (x)¢¢ =6 f (0,1)¢¢ =6 f (y)¢¢ =2x
f (0,1)¢¢ =2(0)
f (0,1)¢¢ =0
f (x)¢y =2y
f (0,1)¢xy =2(1)
f (0,1)¢xy =2
Now find D;
D= f (x)¢ f (y)¢-[ f (x)¢y]2
D=6*0-(2)2
D=0-4
D= -4
As value of D is less than 0 it
will be saddle at (0, 1).
Now checking for points (0, -1);
f (x)¢¢ = 6
f (0,-1)¢¢ = 6
f (y)¢¢ = 2x
f (0,-1)¢¢ = 2x
f (0,-1)¢¢ = 2(0)
f (0,-1)¢¢ = 0
f (x)¢y = 2y
f (0,-1)¢xy = 2y
f (0,-1)¢xy = 2(-1)
f (0,-1)¢xy = -2
Again, finding D;
D= f (x)¢ f (y)¢-[ f (x)¢y]2
D=6*0-(-2)2
D=0-4
D= -4
As also value of D is less than 0
it will be saddle at (0,-1).
Now
checking for points (1/ 6,0);
f (x)¢¢ = 6
f (1/ 6,0)¢¢ = 6
f (y)¢¢ = 2x
f (1/ 6,0)¢¢ = 2x
f (1/ 6,0)¢¢ = 2(1/ 6)
f (1/ 6,0)¢¢ =1/ 3
f (x)¢y = 2y
f (1/ 6,0)¢xy = 2y f (1/ 6,0)¢xy = 2(0) f (1/ 6,0)¢xy = 0
Again, finding D;
D= f (x)¢ f (y)¢-[ f (x)¢y]2
D=6*1/ 3-(0)2
D= 2-0
D= 2
As value of D is greater than 0
it has relative minimum at (1/ 6,0)
By Rehan Ahmed.
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