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

Answer:

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

Step-by-step 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 $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial \lambda} = 0$.

Then, we have

$\frac{\partial F}{\partial x} = 2x-2x\lambda= 2x(1-\lambda)=0$

$\frac{\partial F}{\partial y} = -2y-2y\lambda = -2y(1+\lambda)=0$

$\frac{\partial F}{\partial \lambda} = x^2+y^2-64=0$

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).