Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. (a) f (x, y) = x^2 - y^2; x^2 + y^2 = 1
Max of 1 at (plusminus 1, 0), min of - 1 at (0, plusminus l)
(b) f (x, y) = 3x + y; x^2 + y^2 = 10
Max of 10 at (3, 1), min of - 10 at (- 3, - 1)
(c) f (x, y) = xy; 4x^2 + y^2 = 8
Max of 2 at plusminus (1, 2), min of - 2 at plusminus (l, - 2)

a) f(x,y) = - 1    minimum  at   P ( 0 ; -1 )

b) f (x,y) = 10  maximum at   P ( 3 , 1 ) and   f (x,y) = - 10 minimum at       Q ( - 3 , - 1 )

c)  Max  f ( x , y ) = 2   for points    P ( 1, 2 )  and T ( -1 , -2 )

Min  f ( x , y ) =  -2   for points   Q ( 1 , - 2 )  and R  ( -1 , 2 )

Step-by-step explanation:

A)  f(x,y) = x² - y²         subject to   x² + y² = 1      g(x,y) = x² + y²- 1

δf(x,y)/ δx  = 2*x                                                 δg(x,y)/ δx  =    2*x                      

δf(x,y)/ δy  = - 2*y                                                 δg(x,y)/ δy  =    2*y

δf(x,y)/ δx  = λ* δg(x,y)/ δx

2*x = λ*2*x

δf(x,y)/ δy  = λ* δg(x,y)/ δy

- 2*y = λ*2*y

Then, solving

2*x = λ*2*x           x = λ*x      λ = 1

- 2*y = λ*2*y         y = - 1

x² + y²- 1 = 0         x²  + ( -1)² - 1 = 0      x = 0    

Point  P ( 0 ; -1 )  ; then at that point

f(x,y) = x² - y²             f(x,y) = 0 - ( -1)²     f(x,y) = - 1    minimum

b)  f( x, y ) =  3*x  + y            g ( x , y )  = x² +  y²  = 10

  δf(x,y)/ δx  = 3                   δg(x,y)/ δx  =  2*x

 δf(x,y)/ δy   = 1                   δg(x,y)/ δy  =   2*y

δf(x,y)/ δx  =   λ * δg(x,y)/ δx      ⇒   3 = 2* λ *x    (1)

δf(x,y)/ δy   =  λ *  δg(x,y)/ δy     ⇒    1 = 2*λ * y    (2)

                                                           x² +  y²  - 10 = 0  (3)

Solving that system

From ec (1)     λ  =  3/2*x      From ec (2)     λ  = 1/2*y

Then    (3/2*x )  = 1/2*y          3*y  =  x

x²  +  y²  = 10     ⇒   9y²  + y²  = 10      10*y² = 10

y² = 1       y  ± 1       and    

y  =  1      x  = 3       P  ( 3 , 1  )       y  = - 1    x = -3    Q  ( - 3 , - 1 )

Value of  f( x , y )  at   P   f (x,y) = 3*x + y      f (x,y) =    3*(3) +1  

f (x,y) = 10  maximum at  P ( 3 , 1 )

Value of  f( x , y )  at  Q     f (x,y) = 3*x + y    f (x,y) =  3*(- 3) + ( - 1 )

f (x,y) = - 10 minimum at Q ( - 3 , - 1 )

c)  f( x, y ) =  xy                       g ( x , y )  =  4*x² + y² - 8

δf(x,y)/ δx  = y                         δg(x,y)/ δx  = 8*x

δf(x,y)/ δy = x                          δg(x,y)/ δy  =  2*y

δf(x,y)/ δx  =  λ * δg(x,y)/ δx   ⇒   y  =  λ *8*x    (1)

δf(x,y)/ δy =   λ * δg(x,y)/ δy   ⇒   x  =  λ *2*y    (2)

                                                      4*x² + y² - 8 = 0   (3)

Solving the system

From ec (1)  λ = y/8*x    and From ec (2)      λ = x/2*y        Then    y/8*x  =  x/2*y

2*y² = 8*x²       y² = 4*x²    

Plugging that value in ec (3)

4*x²  +  4*x² - 8 = 0

8*x² = 8     x² = 1        x ± 1      And y² = 4*x²

Then:

for x  = 1      y² =  4     y = ± 2

for x  = -1     y² =  4     y = ± 2

Then we get     P (  1 ; 2 )   Q  ( 1 ; - 2)

                          R ( - 1 ; 2 )  T  ( -1 ; -2)

Plugging that values in f( x , y ) = xy

P (  1 ; 2 )  f( x , y ) = 2            R ( - 1 ; 2 )  f( x , y ) = - 2

Q ( 1 ; - 2)  f( x , y ) = -2           T ( -1 ; -2 )  f( x , y ) = 2

Max  f ( x , y ) = 2   for points    P and T

Min  f ( x , y ) =  -2   for points   Q and R

goog may be able to you find that answer

300pi

step-by-step explanation: