To solve this problem, we’ll use the method of Lagrange multipliers to find the maximum and minimum values of the function $f(x, y) = x^2 - y^2$ subject to the constraint $g(x, y) = x^2 + y^2 - 64 = 0$.

Step 1: Set up the Lagrange function

The method of Lagrange multipliers involves defining a new function $\mathcal{L}(x, y, \lambda)$: $\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \cdot g(x, y)$ So in this case: $\mathcal{L}(x, y, \lambda) = (x^2 - y^2) - \lambda (x^2 + y^2 - 64)$

Step 2: Compute the partial derivatives

We take partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, and $\lambda$, and set them equal to zero.

1. Partial derivative with respect to $x$: $\frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda \cdot 2x = 0$ Simplifying, we get: $x(1 - \lambda) = 0$

2. Partial derivative with respect to $y$: $\frac{\partial \mathcal{L}}{\partial y} = -2y - \lambda \cdot 2y = 0$ Simplifying, we get: $y(-1 - \lambda) = 0$

3. Partial derivative with respect to $\lambda$: $\frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + y^2 - 64) = 0$ which gives us the constraint: $x^2 + y^2 = 64$

Step 3: Solve the equations

From the equations $x(1 - \lambda) = 0$ and $y(-1 - \lambda) = 0$, we have two cases for $\lambda$:

1. Case 1: $x = 0$

   If $x = 0$, then $x^2 + y^2 = 64$ implies $y^2 = 64$, so $y = \pm 8$.

   • For $(x, y) = (0, 8)$, we get $f(0, 8) = 0 - 8^2 = -64$.
   • For $(x, y) = (0, -8)$, we get $f(0, -8) = 0 - (-8)^2 = -64$.

2. Case 2: $y = 0$

   If $y = 0$, then $x^2 + y^2 = 64$ implies $x^2 = 64$, so $x = \pm 8$.

   • For $(x, y) = (8, 0)$, we get $f(8, 0) = 8^2 - 0 = 64$.
   • For $(x, y) = (-8, 0)$, we get $f(-8, 0) = (-8)^2 - 0 = 64$.

Conclusion

The maximum value of $f(x, y)$ is $64$, which occurs at $(x, y) = (8, 0)$ and $(-8, 0)$.

The minimum value of $f(x, y)$ is $-64$, which occurs at $(x, y) = (0, 8)$ and $(0, -8)$.

Final Answer

• Maximum value: $64$
• Minimum value: $-64$

Would you like a more detailed breakdown, or do you have any questions?

Related Questions

1. How would the solution change if the constraint was different, like $x^2 + y^2 = 100$?
2. Can you explain the role of Lagrange multipliers in optimization with constraints?
3. How would you solve this problem if $f(x, y)$ was different, for example $f(x, y) = x^2 + y^2$?
4. What would be the solution if the constraint were an inequality rather than an equation?
5. How do you interpret the physical meaning of this problem if $f(x, y)$ represents height and $x, y$ are coordinates?

Tip

Lagrange multipliers are especially useful in physics and engineering, where constraints like energy conservation often require optimization with multiple variables.