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 = 64

Answer 1

Maximum at points (8,0),(-8,0).Minimum at points (0,8), (0,-8).

Explanation:

There are multiple ways of using lagrange multipliers. Most of them are equivalent.

Consider the function
`F(x,y) = x^2-y^2-\lambda(x^2+y^2-64)`. We want the following
`(\partial F)/(\partial x) = (\partial F)/(\partial y) = (\partial F)/(\partial \lambda) = 0`.

Then, we have

`(\partial F)/(\partial x) = 2x-2x\lambda= 2x(1-\lambda)=0`

`(\partial F)/(\partial y) = -2y-2y\lambda = -2y(1+\lambda)=0`

`(\partial F)/(\partial \lambda) = x^2+y^2-64=0`

From the first two equations, we can see that if
`\lambda =1` then necessarily y=0. IN that case, from the third equation (which is the restriction) gives us that
`x=\pm 8`.

On the other hand, if
`\lambda=-1` then necessarily x=0. Again, using the restriction this gives us that
`y=\pm 8`.

if we evaluate the original function in this points, we have that
`f(0,\pm 8) = -64, f(\pm 8,0)=64`. Then, we have Maximum at points (8,0),(-8,0) and Minimum at points (0,8), (0,-8).