Absolute Minimum 56B8A4

1. **State the problem:** We want to find the absolute minimum value of the function $$f(x, y) = 5 - 2x^2 - y^2$$ within the rectangular domain defined by $$-2 \leq x \leq 2$$ and $$-1 \leq y \leq 1$$. 2. **Understand the function:** The function is a quadratic form with negative coefficients for $x^2$ and $y^2$, which means it opens downward in both $x$ and $y$ directions. The maximum value will be at the vertex, and the minimum will be on the boundary. 3. **Find critical points inside the domain:** Compute partial derivatives: $$\frac{\partial f}{\partial x} = -4x$$ $$\frac{\partial f}{\partial y} = -2y$$ Set them to zero: $$-4x = 0 \Rightarrow x = 0$$ $$-2y = 0 \Rightarrow y = 0$$ So the only critical point inside the domain is at $(0,0)$. 4. **Evaluate $f$ at the critical point:** $$f(0,0) = 5 - 2(0)^2 - (0)^2 = 5$$ 5. **Evaluate $f$ on the boundaries:** - For $x = -2$ or $x = 2$, and $y$ in $[-1,1]$: $$f(\pm 2, y) = 5 - 2(\pm 2)^2 - y^2 = 5 - 2(4) - y^2 = 5 - 8 - y^2 = -3 - y^2$$ Minimum occurs at $y = \pm 1$: $$f(\pm 2, \pm 1) = -3 - 1 = -4$$ - For $y = -1$ or $y = 1$, and $x$ in $[-2,2]$: $$f(x, \pm 1) = 5 - 2x^2 - (\pm 1)^2 = 5 - 2x^2 - 1 = 4 - 2x^2$$ Minimum occurs at $x = \pm 2$: $$f(\pm 2, \pm 1) = 4 - 2(4) = 4 - 8 = -4$$ 6. **Compare values:** - At critical point: $f(0,0) = 5$ - On boundary corners: $f(\pm 2, \pm 1) = -4$ 7. **Conclusion:** The absolute minimum value of $f$ on the domain is $$\boxed{-4}$$, attained at the corners $(\pm 2, \pm 1)$.