Question

examine the extreme values of the function f(x,y)=y2+4xy+3x2+x3​

Answer 1

Answer:

f(x,y) = y² + 4xy + 3x²+ x³

∂f(x,y)/∂x = 0 +  4y  + 6x  + 3x²  

∂f(x,y)/∂y =  2y + 4x + 0 + 0  

∂f(x,y)/∂y = 0 => y = - 2x  

∂f(x,y)/∂x = 0    

substitute  y = - 2x  

4(-2x) +  6x  + 3x²  = 0  

=> 3x² - 2x = 0

=> x(3x - 2) = 0

=> x = 0  , x = 2/3

  y = 0     y = - 4/3

( 0 , 0)  and ( 2/3 , - 4/3)

f(x,y) = y² + 4xy + 3x²+ x³

at ( 0 , 0)

= 0 + 0  + 0 + 0

= 0

f(x,y) = y² + 4xy + 3x²+ x³

( 2/3 , - 4/3)

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

=  16/9  - 32/9  + 12/9  + 8/27

=  -4/9 + 8/27

= -12/27 + 8/27

= -4/27

maximum at (0,0)

minimum at  (23.-4/3)​

Step-by-step explanation:

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