Question

F(x,y)=y²+4xy+3x²+x³

Answer 1

Answer:

Given f(x,y)=x3-4xy+2y2 differentiating equation(i),we have

fx= df/dx = 3 x 2-4y

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

and r = f xx = 6 x

s= f xy = -4

t=f yy=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

Step-by-step explanation:

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