Prob.64. Discuss maxima and minima of the funf(x, y)=x^{3}-4 x y+2 y^{

1.	Prob.64. Discuss maxima and minima of the funf(x, y)=x^{3}-4 x y+2 y^{2}
Answer» Given f(x,y)=x3-4xy+2y2differentiating equation(i),we have fx= df/dx = 3 x2-4y fy= df/dy = -4x+4y and r = fxx= 6 x s= fxy= -4 t=fyy=4 Now for maxima or minima,we must have fx=0,fy=0,we have 3x2-4y=0 (ii) and -4x+4y=0 (iii) Solving equations (ii) and (iii) , we get 3x2-4x=0 or x(3x-4)=0 or x=0, 4/3 Now from equation (iii), we have when x=0 => y=0 and when x=(4/3) => y=(4/3) thus the reuired stationary points are (0,0) and (4/3,4/3) At point (4/3,4/3) r=6 X (4/3) =8 s=-4 t=4 rt-s2=8 X 4 - ( -4)2=32-16=16 > 0 and r > 0 Hence ,f(x,y) has a minima at (4/3,4/3) At point (0,0) r=6 X 0 =0 s=-4 t=4 rt-s2= 0-(-4)2= -16 = -ve and hence there is neither maxima nor minima at (0,0) Minimum value of f(x,y) =[x3- 4xy = 2y2]x = 4/3 , y = 4/3 =(4/3)3-4.(4/3).(4/3)+2.(4/3)2 =(64/27)-(64/9)+(32/9) =(64-192+96)/27 =-(32/27) Ans thank you so much bro

Prob.64. Discuss maxima and minima of the funf(x, y)=x^{3}-4 x y+2 y^{2}

Answer»

Given f(x,y)=x3-4xy+2y2differentiating equation(i),we have

fx= df/dx = 3 x2-4y

fy= df/dy = -4x+4y

and r = fxx= 6 x

s= fxy= -4

t=fyy=4

Now for maxima or minima,we must have fx=0,fy=0,we have

3x2-4y=0 (ii)

and -4x+4y=0 (iii)

Solving equations (ii) and (iii) , we get

3x2-4x=0 or x(3x-4)=0 or x=0, 4/3

Now from equation (iii), we have

when x=0 => y=0 and when x=(4/3) => y=(4/3)

thus the reuired stationary points are (0,0) and (4/3,4/3)

At point (4/3,4/3)

r=6 X (4/3) =8

s=-4

t=4

rt-s2=8 X 4 - ( -4)2=32-16=16 > 0 and r > 0

Hence ,f(x,y) has a minima at (4/3,4/3)

At point (0,0)

r=6 X 0 =0 s=-4

t=4

rt-s2= 0-(-4)2= -16 = -ve

and hence there is neither maxima nor minima at (0,0)

Minimum value of f(x,y) =[x3- 4xy = 2y2]x = 4/3 , y = 4/3

=(4/3)3-4.(4/3).(4/3)+2.(4/3)2

=(64/27)-(64/9)+(32/9)

=(64-192+96)/27

=-(32/27) Ans

thank you so much bro