Question

This extreme value problem 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. f(x, y) = x2 − y2; x2 + y2 = 4

Answer 1

Answer:

• f has a maximum in ( (2,0) , 4) and ( (-2,0) , 4)
• f has a minimum in ( (0,2) , -4 ) and ((0,-2) , -4)

Step-by-step explanation:

f(x,y) = x²-y² and the restriction is

g(x,y) = 4

with g(x,y) = x²+y²

an extreme (x0,y0) of f following that restriction must satisfy

$\nabla f (x_0,y_0) = \lambda \nabla g(x_0,y_0)$

For ceratin constant λ

Now

$\nabla f (x,y) = (2x,-2y)\\\nabla g(x,y) = (2x,2y)$

Thus, they are equal in the first coordinate, so in order for them to be proportional one of the following must happen:

• The second coordinates are equal (this happens only if y=0, because otherwise 2y is not equal to -2y)
• The first coordinate is 0

So, the critical points have the form (x0,0) and (0,y0), also since they have to fulfill the restriction, we have

x0² + 0² = 4, then x0 = 2 or -2

same with (0,y0)

0²+y0² = 4, then y0 must be 2 or -2

Then the critical points are (2,0), (-2,0), (0,2), (0,-2).

In order to find the extremes now we just have to evaluate in each candidate we have and compare the results

f(2,0) = 2² = 4

f(-2,0) = (-2)² = 4

f(0,2) = 0-2² = -4

f(0,-2) = 0-(-2)² = -4

Thus

• f has a maximum in ( (2,0) , 4) and ( (-2,0) , 4)
• f has a minimum in ( (0,2) , -4 ) and ((0,-2) , -4)