To find the absolute maximum and minimum of $f(x, y) = x - y$ on the constraint $x^2 + y^2 \leq 4$, we proceed as follows:

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Step 1: Analyze the interior of the region

The gradient of $f(x, y)$ is: $\nabla f(x, y) = (1, -1).$ Since this gradient does not depend on $x$ or $y$, $f(x, y)$ does not have any critical points in the interior of the disk unless the constraint introduces restrictions. In this case, the boundary is key.

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Step 2: Analyze the boundary $x^2 + y^2 = 4$

On the boundary, we use a parametric representation: $x = 2\cos(t), \quad y = 2\sin(t), \quad t \in [0, 2\pi].$ Substitute into $f(x, y) = x - y$: $f(x, y) = 2\cos(t) - 2\sin(t).$ Let $g(t) = 2(\cos(t) - \sin(t))$. We maximize and minimize $g(t)$ with respect to $t$.

The derivative of $g(t)$ is: $g'(t) = -2\sin(t) - 2\cos(t).$ Set $g'(t) = 0$: $\sin(t) + \cos(t) = 0 \quad \Rightarrow \quad \sin(t) = -\cos(t).$ This implies: $\tan(t) = -1 \quad \Rightarrow \quad t = \frac{3\pi}{4}, \frac{7\pi}{4}.$

Evaluate $g(t)$ at these points and critical angles $t$:

1. At $t = \frac{3\pi}{4}$: $g\left(\frac{3\pi}{4}\right) = 2\left(\cos\left(\frac{3\pi}{4}\right) - \sin\left(\frac{3\pi}{4}\right)\right) = 2\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = -2\sqrt{2}.$

2. At $t = \frac{7\pi}{4}$: $g\left(\frac{7\pi}{4}\right) = 2\left(\cos\left(\frac{7\pi}{4}\right) - \sin\left(\frac{7\pi}{4}\right)\right) = 2\left(\frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right)\right) = 2\sqrt{2}.$

Evaluate $g(t)$ at boundaries $t = 0$ and $t = \pi$:

1. At $t = 0$: $g(0) = 2(\cos(0) - \sin(0)) = 2(1 - 0) = 2.$
2. At $t = \pi$: $g(\pi) = 2(\cos(\pi) - \sin(\pi)) = 2(-1 - 0) = -2.$

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Step 3: Compare results

The extreme values of $f(x, y)$ are:

• Maximum: $2\sqrt{2}$ at $\left(2\cos\frac{7\pi}{4}, 2\sin\frac{7\pi}{4}\right) = \left(\sqrt{2}, -\sqrt{2}\right)$,
• Minimum: $-2\sqrt{2}$ at $\left(2\cos\frac{3\pi}{4}, 2\sin\frac{3\pi}{4}\right) = \left(-\sqrt{2}, \sqrt{2}\right)$.

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Final Answer:

• Absolute Maximum: $f(x, y) = 2\sqrt{2}$.
• Absolute Minimum: $f(x, y) = -2\sqrt{2}$.

Let me know if you'd like more details or visualization!

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Questions for Exploration:

1. How does the gradient $\nabla f(x, y)$ relate to identifying critical points in the interior?
2. What is the role of parametric equations in simplifying boundary analysis?
3. Can this method be adapted for regions bounded by ellipses instead of circles?
4. How would the solution change if the function were $f(x, y) = x + y$?
5. How does symmetry in the domain simplify calculations in problems like these?

Tip:

When optimizing over constrained regions, always verify the boundaries thoroughly, as extremes often occur there.