Question

find the stationary points of the function f(x,y)=x³+y³-3x-12y+20 . Find also the maxima,minima and saddle points.​

Answer 1

SOLUTION

TO DETERMINE

The stationary points of the function f(x,y)=x³+y³-3x-12y+20 .

$\sf{f(x,y) = {x}^{3} + {y}^{3} - 3x - 12y + 20}$

Find also the maxima,minima and saddle points.

EVALUATION

Here the given function is

$\sf{f(x,y) = {x}^{3} + {y}^{3} - 3x - 12y + 20}$

Now

$\sf{f_{x} = 3{x}^{2} - 3}$

$\sf{f_{y} = 3{y}^{2} - 3}$

$\sf{f_{xx} = 6x = A}$

$\sf{f_{yy} = 6y= C}$

$\sf{f_{xy} = 0 = B}$

Now

$\sf{f_{x} = 0 \: \: gives}$

$\sf{ 3{x}^{2} - 3 = 0}$

Which gives x = - 1 , 1

Again

$\sf{f_{y} = 0 \: \: \: gives}$

y = - 2 , 2

Thus there are 4 stationary points and the points are ( - 1 , 2 ), ( - 1 , - 2 ), ( 1, - 2) , ( 1,2)

Now AC - B² = 36xy

For ( - 1 , 2 )

We have AC - B² < 0 & A < 0

So there is neither minimum nor maximum

So ( - 1 , 2 ) is a saddle point

For ( - 1 , - 2 )

We have AC - B² > 0 & A < 0

So there is maximum

So ( - 1 , 2 ) is point with maximum value

For ( 1 , - 2 )

We have AC - B² < 0 & A > 0

So there is neither minimum nor maximum

So ( 1 , - 2 ) is a saddle point

For ( 1 , 2 )

We have AC - B² > 0 & A > 0

So there is a minimum at (1,2)

So ( 1 , 2 ) is a point with minimum value

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