Question

Find the stationary points of the function x3+3xy2-3x2-3y2+4 & also find maximum and minimum values of the function.

Answer 1

Letf`(x,y)=x^3+3xy^2-3x^2-3y^2+4`                    …(1)

∴` f_x=3x^2+3y^2-6x-0+0`

‍∴ `r=f_(x x)=6x-6`                                                  …(2)

Also,` f_y=0+6xy-0-6y+0`

∴` t=f_(y y)=6x-6`                                                      …(3) 

∴` s= f_(x y)=0+6y-0`                                              …(4) 

`"Put" f_x=0 and f_y=0`

∴ `3x^2+3y^2-6x=0`

∴` x^2+y^2-2x=0`                                                  …(5) 

And, `6xy-6y=0` 

∴ `6y(x-1)=0` 

∴ `y=0   pr   x=1`

Case I : When x = 1

From (5), `1^2+y^2-2(1)=0`

∴ `y^2-1=0` 

∴ `y=+-1`

Case II: When y = 0 

from (5),` x^2+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 – s^2 = (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