Exercises — Absolute extrema on closed bounded regions

Type 1 — interior flat spots: $D$ ke andar $\nabla f=0$ solve karo. Yahan $f_x=2x$, $f_y=2y$, toh $\nabla f=0$ sirf $(0,0)$ par hai, jo andar hai. Ek candidate. Type 2 — boundary flat spots: chaar edges mein se har ek par, $f$ ko ek variable mein reduce karo aur dekho kahan uski derivative $0$ hai. (Har edge ek segment hai jaise $x=1,-1\le y\le 1$.) Type 3 — corners: chaar corners $(\pm 1,\pm 1)$, kyunki closed interval par ek 1-variable function apne endpoints par peak kar sakti hai. Hum yahan sirf types list kar rahe hain — abhi koi comparison nahi.

Boundary circle $x^2+y^2=4$ hai. Hum ise Parametrization of curves ka use karke ek variable mein convert karte hain: $x=2\cos t,y=2\sin t,0\le t\le 2\pi .$ Ab fence par $f$ sirf angle $t$ ka function ban jaata hai — ek clean 1-variable problem jisme koi corners nahi hain (circle smooth hai aur khud par close hoti hai, isliye uske "endpoints" $t=0$ aur $t=2\pi$ ek hi point hain).

Interior. $\nabla f=(2x,2y)=0$ at $(0,0)$, andar. $f(0,0)=0$. Boundary. Symmetry se ek edge check karo, maano $x=1,-1\le y\le 1$: $f=1+y^2$, derivative $2y=0$ at $y=0$ giving $f=1$; endpoints $y=\pm 1$ dete hain $f=2$. Chaaon edges par yahi hoga. 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. $\nabla f=(1,2)$ — kabhi bhi zero nahi. Toh koi interior candidates nahi hain: ek plane ke koi flat spots nahi hote. Winners fence par rehte hain. Boundary. Parametrize karo $x=\sqrt{5}\cos t,y=\sqrt{5}\sin t$: $f=\sqrt{5}\cos t+2\sqrt{5}\sin t=\sqrt{5}(\cos t+2\sin t).$ Likho $\cos t+2\sin t=\sqrt{1^2+2^2}\sin (t+\phi )=\sqrt{5}\sin (t+\phi )$ (amplitude trick: $a\cos t+b\sin t$ ki amplitude $\sqrt{a^2+b^2}$ hoti hai). Toh $f=\sqrt{5}\cdot \sqrt{5}\sin (t+\phi )=5\sin (t+\phi )$. Kyunki $\sin$ $[-1,1]$ mein rehta hai: $\boxed{\text{abs max}=5;\text{abs min}=-5.}$ (Yeh length-of-gradient shortcut ke barabar hai: radius $r$ ki disk par linear $f$ ka max $r\Vert \nabla f\Vert =\sqrt{5}\cdot \sqrt{5}=5$ hota hai.)

Interior. $f_x=y=0$ aur $f_y=x=0$ se $(0,0)$ milta hai — yeh ek corner hai, interior nahi. Toh triangle ke strictly andar koi interior critical point nahi hai. Boundary, edge by edge.

• Edge A, $y=0$, $0\le x\le 2$: $f=0$ har jagah.
• Edge B, $x=0$, $0\le y\le 2$: $f=0$ har jagah.
• Edge C, slanted line $(2,0)$ se $(0,2)$ tak: yeh $x+y=2$ satisfy karta hai, toh $y=2-x$, $0\le x\le 2$. Tab $g(x)=x(2-x)=2x-x^2,g^{\prime }(x)=2-2x=0\Rightarrow x=1.$ $x=1$ par: point $(1,1)$, $f=1\cdot 1=1$. Endpoints $x=0,2$ dete hain $f=0$. Corners. $(0,0),(2,0),(0,2)$ sab dete hain $f=0$. Compare. $\{0,1\}$. $\boxed{\text{abs max}=1\text{ at }(1,1);\text{abs min}=0\text{ along both legs}.}$

Interior. $f_x=2x=0,f_y=-2y=0\Rightarrow (0,0)$, andar. $f(0,0)=0$. (Yeh ek saddle hai, lekin humein parwah nahi — hum sirf values compare karte hain.) Boundary. $x=\cos t,y=\sin t$: $f={\cos }^{2}t-{\sin }^{2}t=\cos 2t.$ $\cos 2t$ ki range $[-1,1]$ hai: 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 $(\frac{1}{2},0)$: kya $\frac{1}{4}+0=\frac{1}{4}<1$ hai? Haan, andar. $f(\frac{1}{2},0)=\frac{1}{4}+0-\frac{1}{2}=-\frac{1}{4}$. Boundary $x^2+y^2=1$, toh $y^2=1-x^2$ with $-1\le x\le 1$. Substitute karo: $h(x)=x^2+2(1-x^2)-x=-x^2-x+2.$ $h^{\prime }(x)=-2x-1=0\Rightarrow x=-\frac{1}{2}$. Tab $y^2=1-\frac{1}{4}=\frac{3}{4}$, $y=\pm \frac{\sqrt{3}}{2}$. $h(-\frac{1}{2})=-\frac{1}{4}+\frac{1}{2}+2=\frac{9}{4}$. $x$-range ke endpoints: $x=1$ (point $(1,0)$) deta hai $h=-1-1+2=0$; $x=-1$ (point $(-1,0)$) deta hai $h=-1+1+2=2$. Compare. Candidates: interior $-\frac{1}{4}$; boundary $\frac{9}{4}=2.25$, aur $0$, aur $2$. $\boxed{\text{abs max}=\frac{9}{4}\text{ at }\left(-\frac{1}{2},\pm \frac{\sqrt{3}}{2}\right);\text{abs min}=-\frac{1}{4}\text{ at }\left(\frac{1}{2},0\right).}$ $y^2=1-x^2$ substitute kyun kiya (parametrize kyun nahi): kyunki $f$ mein $y$ sirf $y^2$ ke roop mein aata hai — substitution turant $y$ ko bina kisi trig ke hata deta hai. Woh tool chunno jo function ke saath match kare.

Lagrange. Hum chahte hain $\nabla f=\lambda \nabla g$ jahan $g=x^2+y^2-8$ hai. $f_x=y=\lambda 2x,f_y=x=\lambda 2y.$ Multiply karo: $y\cdot x=\lambda 2x\cdot ?$ — dono equations divide karna zyada clean hai. Unse, $y=2\lambda x$ aur $x=2\lambda y$, toh $x=2\lambda (2\lambda x)=4{\lambda }^{2}x$. Agar $x\ne 0$: $4{\lambda }^2=1$, $\lambda =\pm \frac{1}{2}$.

• $\lambda =\frac{1}{2}$: $y=x$. $x^2+y^2=8$ par: $2x^2=8$, $x=\pm 2$, points $(2,2),(-2,-2)$, $f=4$.
• $\lambda =-\frac{1}{2}$: $y=-x$. Tab $2x^2=8$, $x=\pm 2$, points $(2,-2),(-2,2)$, $f=-4$. Parametrize karke confirm karo. $x=2\sqrt{2}\cos t,y=2\sqrt{2}\sin t$: $f=8\cos t\sin t=4\sin 2t\in [-4,4].$ $\boxed{\text{abs max}=4\text{ at }(\pm 2,\pm 2)(\text{same signs});\text{abs min}=-4\text{ at }(\pm 2,\mp 2).}$ Dono methods agree karte hain.

Interior. $f_x=2x-2=0\Rightarrow x=1$; $f_y=2y-2=0\Rightarrow y=1$. Point $(1,1)$: kya yeh triangle $x\ge 0,y\ge 0,x+y\le 4$ ke andar hai? $1+1=2\le 4$ ✓. $f(1,1)=1+1-2-2=-2$. 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, $x=0$, $0\le y\le 4$: symmetry se $f=y^2-2y$, 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. Product rule use karo. Maano $r^2=x^2+y^2$. $f_x=e^{-r^2}(1-2x^2),f_y=x\cdot e^{-r^2}(-2y)=-2xy{e}^{-r^2}.$ $e^{-r^2}>0$ hamesha, toh yeh kabhi zero cause nahi karta. Dono brackets ko $0$ set karo:

• $f_y=0$: $xy=0$, toh $x=0$ ya $y=0$.
• Agar $x=0$: tab $f_x=e^{-y^2}(1-0)=e^{-y^2}\ne 0$. Koi solution nahi.
• Agar $y=0$: tab $f_x=e^{-x^2}(1-2x^2)=0\Rightarrow x^2=\frac{1}{2},x=\pm \frac{1}{\sqrt{2}}$. Points $(\pm \frac{1}{\sqrt{2}},0)$: radius${}^2=\frac{1}{2}<1$, dono andar. $f(\frac{1}{\sqrt{2}},0)=\frac{1}{\sqrt{2}}{e}^{-1/2}$; $f(-\frac{1}{\sqrt{2}},0)=-\frac{1}{\sqrt{2}}{e}^{-1/2}$. Numerically $\frac{1}{\sqrt{2}}{e}^{-1/2}\approx 0.7071\cdot 0.6065\approx 0.4289$. Boundary. $x^2+y^2=1$ par, $e^{-r^2}=e^{-1}$ constant hai, toh $f=x{e}^{-1}$. Parametrize karo $x=\cos t$: $f=e^{-1}\cos t\in [-e^{-1},e^{-1}]$. Toh boundary max $=e^{-1}\approx 0.3679$, min $=-e^{-1}\approx -0.3679$. Compare. Interior $\pm 0.4289$ boundary $\pm 0.3679$ se beat karta hai. $\boxed{\text{abs max}=\frac{1}{\sqrt{2}}{e}^{-1/2}\approx 0.4289\text{ at }(\frac{1}{\sqrt{2}},0);\text{abs min}=-\frac{1}{\sqrt{2}}{e}^{-1/2}\approx -0.4289\text{ at }(-\frac{1}{\sqrt{2}},0).}$ Exponential factor critical point kyun nahi deta: $e^{\text{kuch bhi}}$ strictly positive hota hai, toh yeh kabhi zero nahi ho sakta. Yeh sirf $f$ ko scale karta hai — flat spots poori tarah polynomial-and-derivative brackets se aate hain. "Kaun sa factor zero ho sakta hai" ko pehchanna mastery move hai.