Find all the local maxima local minima and saddle points of

Find all the local maxima, local minima, and saddle points of the function f(x y)-x-4xyy+ 12y +2 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice O A. A local maximum occurs at 1 (Type an ordered pair. Use a comma to separate answers as needed) The loal maximum value(s) is/are (Type an exact answer. Use a comma to separate answers as needed) O B. There are no local maxima

Solution

1 ) Given that

f(x,y) = x² - 4xy + y² +12y + 2

f_x = f/x = 2x - 4y

f_y = f /y = -4x +2y + 12

For finding stationary points ,

f_x=0 , f_y = 0

2x - 4y = 0

2x = 4y ....................................................1

-4x + 2y = -12

2( -2x + y ) = -2.6

-2x + y = -6

2x = y + 6.........................................2

Equating the equations 1 and 2

4y = y + 6

4y - y = 6

3y = 6

y = 6/3

y = 2

Substitute y = 2 in equation 1

2x = 4y

2x = 4(2)

2x = 8

x = 8/2

x = 4

Therefore,

Stationary point is , ( x , y ) = ( 4 , 2 )

f_xx = ²f / x²

= /x(2x - 4y)

= 2

f_yy = ²f/y²

= /y(-4x +2y + 12)

= 2

f_xy = ²f / xy = -4

We compute,

f_xx.f_yy - f²_xy = 2.2 - (-4)²

= 4 - 16

= -12 < 0

f_xx.f_yy - f²_xy < 0 then the stationary point is saddle point

Hence, (4 , 2 ) is a saddle point

2 ) Given that

f(x,y) = 2x² + 3xy + 4y² - 7x - 11y

f_x = f/x = 4x + 3y - 7

f_y = f /y = 3x + 8y - 11

For finding stationary points ,

    f_x=0 , f_y = 0

4x + 3y - 7 = 0

4x + 3y = 7..........................1

3x + 8y - 11 = 0

3x + 8y = 11..............................2

On solving equations 1 and 2

( x , y ) = ( - 89/23 , 65/23)

We compute,

f_xx = ²f / x²  

= /x( 4x + 3y - 7 )

= 4

f_yy = ²f/y²

= /y(3x + 8y - 11)

= 8

f_xy = ²f / xy = 3

We compute,

f_xx.f_yy - f²_xy = 4.8 - (3)²

= 32 - 9

= 23 > 0

f_xx._fyy - f²_xy > 0 then the stationary point is either local maximum or local minimum

   Since f_xx < 0 and f_yy < 0 it is a local maximum.

   f_xx > 0 and f_yy > 0 it is a local miniimum.

Here,

f_xx = 4 > 0

f_xx > 0

f_yy = 8 > 0

f_yy > 0

Hence,

f_xx > 0 , f_yy > 0 then the stationary point ( x , y ) = ( - 89/23 , 65/23) is local minimum

Therefore,

The function is local minimum at (  - 89/23 , 65/23)

3 ) Given that

f(x,y) = 15x² -2x³ +3y² + 6xy

f_x = f/x = 30x - 6x + 6y = 24x + 6y

f_y = f /y = 6y + 6x = 6x + 6y

For finding stationary points ,

    f_x=0 , f_y = 0

24x + 6y = 0

6y = -24x..........................1

6x + 6y = 0

6y = -6x..............................2

Solving equations 1 and 2

Equating the equations 1 and 2

-24x = -6x

-24x + 6x = 0

-18x = 0

x = 0

Substitute x = 0 in equation 1

6y = -24x

6y = -24(0)

6y = 0

y = 0

( x , y ) = ( 0 , 0)

We compute,

f_xx = ²f / x²  

= /x( 24x + 6y )

= 24

f_yy = ²f/y²

= /y( 6x + 6y)

= 6

f_xy = ²f / xy = 6

We compute,

f_xx.f_yy - f²_xy = 24.6 - (6)²

= 144 - 36

= 108 > 0

f_xx._fyy - f²_xy > 0 then the stationary point is either local maximum or local minimum

   Since f_xx < 0 and f_yy < 0 it is a local maximum.

   f_xx > 0 and f_yy > 0 it is a local miniimum.

Here,

f_xx = 24 > 0

f_xx > 0

f_yy = 6 > 0

f_yy > 0

Hence,

f_xx > 0 , f_yy > 0 then the stationary point ( x , y ) = ( 0 , 0) is local minimum

Therefore,

The function is local minimum at ( 0 , 0 )

4 ) Given that

f(x , y ) = x² + xy + y² +4x - 4y + 6

f_x = f/x = 2x + y + 4

f_y = f /y = x + 2y - 4

For finding stationary points ,

    f_x=0 , f_y = 0

2x + y + 4= 0

2x + y = - 4..........................1

x + 2y - 4 = 0

x + 2y = 4..............................2

Solving equations 1 and 2

Multiply equation 2 with 2

x + 2y = 4

2x + 4y = 8

2x = 8 - 4y..................................3

From equation 1

2x + y = -4

2x = - 4 - y...............................4

Solve equations 3 and 4

2x = 8 - 4y

2x = - 4 - y

Equating the equations 3 and 4

8 - 4y = - 4 - y

-4y + y = -4 - 8

-3y = -12

y = 4

Substitute y = 4 in equation 3

2x = 8 - 4y

2x = 8 - 4(3)

2x = 8 - 12

2x = -4

x = -2

( x , y ) = ( - 2 , 3 )

We compute,

f_xx = ²f / x²  

= /x( 2x+y+4 )

= 2

f_yy = ²f/y²

= /y(x+2y-4)

= 2

f_xy = ²f / xy = 1

We compute,

f_xx.f_yy - f²_xy = 2.2 - (1)²

= 4 - 1

= 3 > 0

f_xx._fyy - f²_xy > 0 then the stationary point is either local maximum or local minimum

   Since f_xx < 0 and f_yy < 0 it is a local maximum.

   f_xx > 0 and f_yy > 0 it is a local miniimum.

Here,

f_xx = 2 > 0

f_xx > 0

f_yy = 2 > 0

f_yy > 0

Hence,

f_xx > 0 , f_yy > 0 then the stationary point ( x , y ) = ( -2 , 3) is local minimum

Therefore,

The function is local minimum at ( -2 , 3 )