We are tasked with finding the maximum and minimum values of the function:

$f(x, y) = 3^2x^2 - 2^2y^2 = 9x^2 - 4y^2$

on the rectangular region defined by $-2 \leq x \leq 2$ and $-3 \leq y \leq 3$. To do this, we will follow these steps:

Step 1: Compute the Gradient

To find the critical points, we need to compute the gradient of the function $f(x, y)$. The gradient is given by:

$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

Partial Derivative with Respect to $x$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 9x^2 - 4y^2 \right) = 18x$

Partial Derivative with Respect to $y$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( 9x^2 - 4y^2 \right) = -8y$

Thus, the gradient is: $\nabla f(x, y) = (18x, -8y)$

Step 2: Set the Gradient Equal to Zero

To find critical points, we set $\nabla f(x, y) = 0$:

$18x = 0 \quad \text{and} \quad -8y = 0$

Solving these, we get: $x = 0 \quad \text{and} \quad y = 0$

Thus, the only critical point in the region is $(0, 0)$.

Step 3: Evaluate the Function at the Critical Point

We evaluate $f(x, y)$ at $(0, 0)$:

$f(0, 0) = 9(0)^2 - 4(0)^2 = 0$

So, $f(0, 0) = 0$.

Step 4: Check the Boundaries of the Region

Next, we evaluate $f(x, y)$ at the boundaries of the rectangular region.

On the boundary $x = 2$:

$f(2, y) = 9(2)^2 - 4y^2 = 36 - 4y^2$ We check the values for $y = -3$ and $y = 3$:

$f(2, 3) = 36 - 4(3)^2 = 36 - 36 = 0$ $f(2, -3) = 36 - 4(-3)^2 = 36 - 36 = 0$

Similarly, for $x = -2$: $f(-2, 3) = 36 - 36 = 0 \quad \text{and} \quad f(-2, -3) = 36 - 36 = 0$

On the boundary $y = 3$:

$f(x, 3) = 9x^2 - 4(3)^2 = 9x^2 - 36$ We check the values for $x = 2$ and $x = -2$:

$f(2, 3) = 36 - 36 = 0 \quad \text{and} \quad f(-2, 3) = 36 - 36 = 0$

Similarly, for $y = -3$, we get: $f(x, -3) = 9x^2 - 36$ And at $x = 2$ and $x = -2$: $f(2, -3) = 0 \quad \text{and} \quad f(-2, -3) = 0$

On the boundary $x = 0$:

$f(0, y) = 9(0)^2 - 4y^2 = -4y^2$ We check the values for $y = -3$ and $y = 3$:

$f(0, 3) = -36 \quad \text{and} \quad f(0, -3) = -36$

Thus, the function takes its minimum value of $-36$ at $(0, 3)$ and $(0, -3)$.

Step 5: Conclusion

The function $f(x, y) = 9x^2 - 4y^2$ achieves its maximum value of 36 at points $(2, 0)$ and $(-2, 0)$, and its minimum value of -36 at points $(0, 3)$ and $(0, -3)$.

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Would you like more details on any of the steps? Here are 5 related questions you can explore:

1. How would the results change if the region were circular instead of rectangular?
2. What do the level curves of this function represent geometrically?
3. How does the choice of constants (like 9 and -4) affect the shape of the function?
4. Can you apply this method to more complex functions like trigonometric functions?
5. How do we determine the nature of critical points using the second derivative test?

Tip: Always check the boundaries when finding extrema in constrained regions, as the maximum or minimum might occur there.