Exercises — Absolute extrema on closed bounded regions

Type 1 — interior flat spots: solve $\nabla f=0$ inside $D$. Here $f_x=2x$, $f_y=2y$, so $\nabla f=0$ only at $(0,0)$, which is inside. One candidate. Type 2 — boundary flat spots: on each of the four edges, reduce $f$ to one variable and find where its derivative is $0$. (Each edge is a segment like $x=1,-1\le y\le 1$.) Type 3 — corners: the four corners $(\pm 1,\pm 1)$, because a 1-variable function on a closed interval can peak at its endpoints. We are only listing the types here — no comparison yet.

The boundary is the circle $x^2+y^2=4$. We turn it into one variable using Parametrization of curves: $x=2\cos t,y=2\sin t,0\le t\le 2\pi .$ Now $f$ on the fence becomes a function of the single angle $t$ — a clean 1-variable problem with no corners (a circle is smooth and closes on itself, so its "endpoints" $t=0$ and $t=2\pi$ are the same point).

Interior. $\nabla f=(2x,2y)=0$ at $(0,0)$, inside. $f(0,0)=0$. Boundary. By symmetry check one edge, say $x=1,-1\le y\le 1$: $f=1+y^2$, derivative $2y=0$ at $y=0$ giving $f=1$; endpoints $y=\pm 1$ give $f=2$. Same story on all four edges. Corners. $(\pm 1,\pm 1)$: $f=1+1=2$. Compare. Values collected: $0,1,2$. $\boxed{\text{abs max}=2\text{ at the four corners};\text{abs min}=0\text{ at }(0,0).}$

Interior. $\nabla f=(1,2)$ — never zero. So there are no interior candidates: a plane has no flat spots. The winners must live on the fence. Boundary. Parametrize $x=\sqrt{5}\cos t,y=\sqrt{5}\sin t$: $f=\sqrt{5}\cos t+2\sqrt{5}\sin t=\sqrt{5}(\cos t+2\sin t).$ Write $\cos t+2\sin t=\sqrt{1^2+2^2}\sin (t+\phi )=\sqrt{5}\sin (t+\phi )$ (amplitude trick: $a\cos t+b\sin t$ has amplitude $\sqrt{a^2+b^2}$). So $f=\sqrt{5}\cdot \sqrt{5}\sin (t+\phi )=5\sin (t+\phi )$. Since $\sin$ lives in $[-1,1]$: $\boxed{\text{abs max}=5;\text{abs min}=-5.}$ (This equals the length-of-gradient shortcut: max of a linear $f$ on a disk of radius $r$ is $r\Vert \nabla f\Vert =\sqrt{5}\cdot \sqrt{5}=5$.)

Interior. $f_x=y=0$ and $f_y=x=0$ force $(0,0)$ — a corner, not interior. So there is no interior critical point strictly inside the triangle. Boundary, edge by edge.

• Edge A, $y=0$, $0\le x\le 2$: $f=0$ everywhere.
• Edge B, $x=0$, $0\le y\le 2$: $f=0$ everywhere.
• Edge C, the slanted line from $(2,0)$ to $(0,2)$: it satisfies $x+y=2$, so $y=2-x$, $0\le x\le 2$. Then $g(x)=x(2-x)=2x-x^2,g^{\prime }(x)=2-2x=0\Rightarrow x=1.$ At $x=1$: point $(1,1)$, $f=1\cdot 1=1$. Endpoints $x=0,2$ give $f=0$. Corners. $(0,0),(2,0),(0,2)$ all give $f=0$. Compare. $\{0,1\}$. $\boxed{\text{abs max}=1\text{ at }(1,1);\text{abs min}=0\text{ along both legs}.}$

Interior. $f_x=2x=0,f_y=-2y=0\Rightarrow (0,0)$, inside. $f(0,0)=0$. (This is a saddle, but we don't care — we only compare values.) Boundary. $x=\cos t,y=\sin t$: $f={\cos }^{2}t-{\sin }^{2}t=\cos 2t.$ $\cos 2t$ ranges over $[-1,1]$: max $1$ at $t=0,\pi$ (points $(\pm 1,0)$), min $-1$ at $t=\frac{\pi }{2},\frac{3\pi }{2}$ (points $(0,\pm 1)$). Compare. Interior $0$ vs boundary $\pm 1$. $\boxed{\text{abs max}=1\text{ at }(\pm 1,0);\text{abs min}=-1\text{ at }(0,\pm 1).}$

Interior. $f_x=2x-1=0\Rightarrow x=\frac{1}{2}$; $f_y=4y=0\Rightarrow y=0$. Point $(\frac{1}{2},0)$: is $\frac{1}{4}+0=\frac{1}{4}<1$? Yes, inside. $f(\frac{1}{2},0)=\frac{1}{4}+0-\frac{1}{2}=-\frac{1}{4}$. Boundary $x^2+y^2=1$, so $y^2=1-x^2$ with $-1\le x\le 1$. Substitute: $h(x)=x^2+2(1-x^2)-x=-x^2-x+2.$ $h^{\prime }(x)=-2x-1=0\Rightarrow x=-\frac{1}{2}$. Then $y^2=1-\frac{1}{4}=\frac{3}{4}$, $y=\pm \frac{\sqrt{3}}{2}$. $h(-\frac{1}{2})=-\frac{1}{4}+\frac{1}{2}+2=\frac{9}{4}$. Endpoints of the $x$-range: $x=1$ (point $(1,0)$) gives $h=-1-1+2=0$; $x=-1$ (point $(-1,0)$) gives $h=-1+1+2=2$. Compare. Candidates: interior $-\frac{1}{4}$; boundary $\frac{9}{4}=2.25$, and $0$, and $2$. $\boxed{\text{abs max}=\frac{9}{4}\text{ at }\left(-\frac{1}{2},\pm \frac{\sqrt{3}}{2}\right);\text{abs min}=-\frac{1}{4}\text{ at }\left(\frac{1}{2},0\right).}$ Why substitute $y^2=1-x^2$ (not parametrize)? Because $f$ contains $y$ only as $y^2$ — the substitution instantly removes $y$ with no trig at all. Choose the tool that matches the function.

Lagrange. We want $\nabla f=\lambda \nabla g$ with $g=x^2+y^2-8$. $f_x=y=\lambda 2x,f_y=x=\lambda 2y.$ Multiply: $y\cdot x=\lambda 2x\cdot ?$ — cleaner to divide the two equations. From them, $y=2\lambda x$ and $x=2\lambda y$, so $x=2\lambda (2\lambda x)=4{\lambda }^{2}x$. If $x\ne 0$: $4{\lambda }^2=1$, $\lambda =\pm \frac{1}{2}$.

• $\lambda =\frac{1}{2}$: $y=x$. On $x^2+y^2=8$: $2x^2=8$, $x=\pm 2$, points $(2,2),(-2,-2)$, $f=4$.
• $\lambda =-\frac{1}{2}$: $y=-x$. Then $2x^2=8$, $x=\pm 2$, points $(2,-2),(-2,2)$, $f=-4$. Confirm by parametrizing. $x=2\sqrt{2}\cos t,y=2\sqrt{2}\sin t$: $f=8\cos t\sin t=4\sin 2t\in [-4,4].$ $\boxed{\text{abs max}=4\text{ at }(\pm 2,\pm 2)(\text{same signs});\text{abs min}=-4\text{ at }(\pm 2,\mp 2).}$ Both methods agree.

Interior. $f_x=2x-2=0\Rightarrow x=1$; $f_y=2y-2=0\Rightarrow y=1$. Point $(1,1)$: is it inside the triangle $x\ge 0,y\ge 0,x+y\le 4$? $1+1=2\le 4$ ✓. $f(1,1)=1+1-2-2=-2$. Boundary.

• Edge A, $y=0$, $0\le x\le 4$: $f=x^2-2x$, $f^{\prime }=2x-2=0\Rightarrow x=1$, $f(1,0)=1-2=-1$. Ends: $f(0,0)=0$, $f(4,0)=16-8=8$.
• Edge B, $x=0$, $0\le y\le 4$: by symmetry $f=y^2-2y$, crit $y=1$: $f(0,1)=-1$. Ends: $f(0,0)=0$, $f(0,4)=8$.
• Edge C, $x+y=4\Rightarrow y=4-x$, $0\le x\le 4$: $f=x^2+(4-x{)}^2-2x-2(4-x)=2x^2-8x+8.$ $f^{\prime }=4x-8=0\Rightarrow x=2$, point $(2,2)$: $f=8-16+8=0$. Ends already counted. Corners. $(0,0)=0$; $(4,0)=8$; $(0,4)=8$. Compare. $\{-2,-1,0,8\}$. $\boxed{\text{abs max}=8\text{ at }(4,0)\text{ and }(0,4);\text{abs min}=-2\text{ at }(1,1).}$

Interior. Use the product rule. Let $r^2=x^2+y^2$. $f_x=e^{-r^2}(1-2x^2),f_y=x\cdot e^{-r^2}(-2y)=-2xy{e}^{-r^2}.$ $e^{-r^2}>0$ always, so it never causes a zero. Set both brackets to $0$:

• $f_y=0$: $xy=0$, so $x=0$ or $y=0$.
• If $x=0$: then $f_x=e^{-y^2}(1-0)=e^{-y^2}\ne 0$. No solution.
• If $y=0$: then $f_x=e^{-x^2}(1-2x^2)=0\Rightarrow x^2=\frac{1}{2},x=\pm \frac{1}{\sqrt{2}}$. Points $(\pm \frac{1}{\sqrt{2}},0)$: radius${}^2=\frac{1}{2}<1$, both inside. $f(\frac{1}{\sqrt{2}},0)=\frac{1}{\sqrt{2}}{e}^{-1/2}$; $f(-\frac{1}{\sqrt{2}},0)=-\frac{1}{\sqrt{2}}{e}^{-1/2}$. Numerically $\frac{1}{\sqrt{2}}{e}^{-1/2}\approx 0.7071\cdot 0.6065\approx 0.4289$. Boundary. On $x^2+y^2=1$, $e^{-r^2}=e^{-1}$ is constant, so $f=x{e}^{-1}$. Parametrize $x=\cos t$: $f=e^{-1}\cos t\in [-e^{-1},e^{-1}]$. So boundary max $=e^{-1}\approx 0.3679$, min $=-e^{-1}\approx -0.3679$. Compare. Interior $\pm 0.4289$ beat boundary $\pm 0.3679$. $\boxed{\text{abs max}=\frac{1}{\sqrt{2}}{e}^{-1/2}\approx 0.4289\text{ at }(\frac{1}{\sqrt{2}},0);\text{abs min}=-\frac{1}{\sqrt{2}}{e}^{-1/2}\approx -0.4289\text{ at }(-\frac{1}{\sqrt{2}},0).}$ Why the exponential factor never gives a critical point by itself: $e^{\text{anything}}$ is strictly positive, so it can never equal zero. It only scales $f$ — the flat spots come entirely from the polynomial-and-derivative brackets. Recognising "which factor can be zero" is the mastery move.