If f(x, y) = 3x2 – 3y2 + x3 then the value stationary point of f(x,y)are
Answers
Answer:
Letf
(x,y)=x3+3xy2-3x2-3y2+4 …(1)
∴
fx=3x2+3y2-6x-0+0
∴
r=fxx=6x-6 …(2)
Also,
fy=0+6xy-0-6y+0
∴
t=fyy=6x-6 …(3)
∴
s=fxy=0+6y-0 …(4)
Putfx=0andfy=0
∴
3x2+3y2-6x=0
∴
x2+y2-2x=0 …(5)
And, 6xy-6y=0
∴ 6y(x-1)=0
∴ y=0 pr x=1
Case I : When x = 1
From (5),
12+y2-2(1)=0
∴
y2-1=0
∴ y=±1
Case II: When y = 0
from (5),
x2+0-2x=0
∴ x(x-2)=0
∴x=0 or x=2
∴ Stationary points are (1,1);(1,-1);(0,0);(2,0):
(i)At (1,1)
From (2), r = 6(1) – 6 = 0
∴ f is neither maximum or minimum at (1,1)
(ii)At (1,-1)
From (2), r = 6(1) – 6 = 0
∴ f is neither maximum or minimum at (1,-1)
(iii)At (0,0)
From (2), r = 6(0) – 6 = - 6 < 0
From (3), t = 6(0) – 6 = - 6
From (4), s = 6(0) = 0
∴ rt – s2 = (-6)(-6) – 0 = 36 > 0
∴ f has maximum at (0,0)
From (1), Maximum value of f
∴f = (0)3 + 3(0)(0)2 – 3(0)2 – 3(0)2 + 4 = 4
(iv)At (2,0)
From (2), r = 6(2) – 6 = 6 < 0
From (3), t = 6(2) – 6 = 6
From (4), s = 6(0) = 0
∴
rt–s2=(6)(6)–0=36>0`
∴" f has maximum at (2,0)
From (1), Minimum value of f"
f=(2)3+3(2)(0)2-3(2)0-3(2)2-3(0)2+4=0
Hence the function has
Maximum at (0,0) and Maximum value = 4
Minimum at (2,0) and Minimum value = 0