Question

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)

Answer 1

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 )

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