Absolute Maximum & Minimum on a Region

$f_x = 4x - 4 = 0$, $f_y = 2y + 2 = 0$ give $(1, -1)$, which is inside. $f(1, -1) = -3$.

$x = 0$: $y^2 + 2y$. $\ x = 3$: $y^2 + 2y + 6$. $\ y = -2$: $2x^2 - 4x$. $\ y = 1$: $2x^2 - 4x + 3$.

Vertical edges: $2y + 2 = 0$ at $y = -1$, so $f(0, -1) = -1$ and $f(3, -1) = 5$. Horizontal edges: $4x - 4 = 0$ at $x = 1$, so $f(1, -2) = -2$ and $f(1, 1) = 1$.

$f(0, -2) = 0$, $f(3, -2) = 6$, $f(0, 1) = 3$, $f(3, 1) = 9$.

Candidates: $-3, -1, 5, -2, 1, 0, 6, 3, 9$. Largest $9$, smallest $-3$.

$f_x = y - 1 = 0$, $f_y = x - 2 = 0$ give $(2, 1)$, inside. $f(2, 1) = -2$.

$x = 0$: $-2y$. $\ x = 4$: $2y - 4$. $\ y = 0$: $-x$. $\ y = 3$: $2x - 6$.

All four are lines with nonzero slope: no critical points on the edges.

$f(0, 0) = 0$, $f(4, 0) = -4$, $f(0, 3) = -6$, $f(4, 3) = 2$.

Candidates: $-2, 0, -4, -6, 2$. The interior point ($-2$) is neither the max nor the min.

$f_x = 2x = 0$, $f_y = 4y - 2 = 0$ give $\left(0, \tfrac12\right)$, inside. $f\left(0, \tfrac12\right) = -\tfrac12$.

On the circle $x^2 = 4 - y^2$: $g(y) = 4 - y^2 + 2y^2 - 2y = y^2 - 2y + 4$, for $-2 \le y \le 2$.

$g'(y) = 2y - 2 = 0$ at $y = 1$, $x = \pm\sqrt3$: $g = 3$. Endpoints: $g(2) = 4$ at $(0, 2)$, $g(-2) = 12$ at $(0, -2)$.

Candidates: $-\tfrac12, 3, 4, 12$.

$f_x = 1 \neq 0$: no critical points inside.

$x = \sqrt5\cos t$, $y = \sqrt5\sin t$: $g(t) = \sqrt5\cos t - 2\sqrt5\sin t + 1$.

$g'(t) = -\sqrt5\sin t - 2\sqrt5\cos t = 0$ gives $\tan t = -2$, so $(\cos t, \sin t) = \pm\left(\tfrac{1}{\sqrt5}, -\tfrac{2}{\sqrt5}\right)$: the points $(1, -2)$ and $(-1, 2)$. A full circle has no endpoints.

$f(1, -2) = 6$ and $f(-1, 2) = -4$.

$T_x = 2x = 0$, $T_y = 2y - 2 = 0$ give $(0, 1)$, inside. $T(0, 1) = -1$.

Arc: $x^2 = 9 - y^2$, so $g(y) = 9 - 2y$ for $0 \le y \le 3$. Flat edge $y = 0$: $h(x) = x^2$ for $-3 \le x \le 3$.

$g' = -2 \neq 0$, so only its endpoints: $T(\pm 3, 0) = 9$, $T(0, 3) = 3$. $\ h' = 2x = 0$ at $(0, 0)$: $T = 0$.

Candidates: $-1, 9, 3, 0$.

Edges $x = 0$ ($0 \le y \le 4$), $y = 0$ ($0 \le x \le 4$) and $y = 4 - x$ ($0 \le x \le 4$). Corners $(0, 0)$, $(4, 0)$, $(0, 4)$.

$f_x = 2x - 2 = 0$, $f_y = 2y - 2 = 0$ give $(1, 1)$, inside. $f(1, 1) = -2$.

$x = 0$: $y^2 - 2y$. $\ y = 0$: $x^2 - 2x$. $\ y = 4 - x$: $x^2 + (4 - x)^2 - 2x - 2(4 - x) = 2x^2 - 8x + 8$.

$(0, 1)$: $-1$. $\ (1, 0)$: $-1$. $\ 4x - 8 = 0$ at $(2, 2)$: $0$. Corners: $f(0, 0) = 0$, $f(4, 0) = 8$, $f(0, 4) = 8$.

Candidates: $-2, -1, -1, 0, 0, 8, 8$.

Bottom $y = x^2$, top $y = 4$, $-2 \le x \le 2$; corners $(\pm 2, 4)$.

$f_x = 2x = 0$, $f_y = 4y - 4 = 0$ give $(0, 1)$; $0 < 1 < 4$, so inside. $f(0, 1) = -2$.

Top: $g(x) = x^2 + 16$. Bottom: $h(x) = x^2 + 2x^4 - 4x^2 = 2x^4 - 3x^2$.

$g' = 2x = 0$ at $(0, 4)$: $16$. $\ h' = 2x(4x^2 - 3) = 0$ at $(0, 0)$: $0$, and at $\left(\pm\tfrac{\sqrt3}{2}, \tfrac34\right)$: $-\tfrac98$. Corners $(\pm 2, 4)$: $20$.

Candidates: $-2, 16, 0, -\tfrac98, 20$.

$f_x = y = 0$, $f_y = x = 0$ give $(0, 0)$, inside. $f(0, 0) = 0$.

$y = 2\lambda x$, $\ x = 8\lambda y$. Multiply the first by $x$ and the second by $y$: $xy = 2\lambda x^2$ and $xy = 8\lambda y^2$. If $\lambda = 0$, then $x = y = 0$, which is not on the curve. So $x^2 = 4y^2$, and $8y^2 = 8$: $y = \pm 1$, $x = \pm 2$.

$f(2, 1) = f(-2, -1) = 2$; $\ f(2, -1) = f(-2, 1) = -2$; $\ f(0, 0) = 0$.

$f_x = 2x - 1 = 0$, $f_y = 4y = 0$ give $\left(\tfrac12, 0\right)$, inside. $f = -\tfrac14$.

$2x - 1 = 2\lambda x$, $\ 4y = 2\lambda y$. Second: $y = 0$ or $\lambda = 2$. $y = 0$: $(\pm 1, 0)$. $\lambda = 2$: $2x - 1 = 4x$, $x = -\tfrac12$, $y = \pm\tfrac{\sqrt3}{2}$.

$f(1, 0) = 0$, $\ f(-1, 0) = 2$, $\ f\left(-\tfrac12, \pm\tfrac{\sqrt3}{2}\right) = \tfrac14 + \tfrac32 + \tfrac12 = \tfrac94$, $\ f\left(\tfrac12, 0\right) = -\tfrac14$.