📝 Çözümlü Örnekler

Çözüm: $L(x,y,\lambda)=x^2+y^2-\lambda(x+y-1)$
FOC:
$L_x=2x-\lambda=0 \Rightarrow \lambda=2x$
$L_y=2y-\lambda=0 \Rightarrow \lambda=2y$
$L_\lambda=-(x+y-1)=0 \Rightarrow x+y=1$
$2x=2y \Rightarrow x=y$
$2x=1 \Rightarrow x=y=0.5$
$\boxed{x=0.5,\ y=0.5}$ noktasında $f_{\text{min}}=0.5$

Çözüm: $L=xy-\lambda(x+y-4)$
$L_x=y-\lambda=0 \Rightarrow \lambda=y$
$L_y=x-\lambda=0 \Rightarrow \lambda=x$
$x=y$ ve $x+y=4 \Rightarrow x=y=2$
$\boxed{x=2,\ y=2}$ noktasında $f_{\text{max}}=4$

Çözüm: $L=xyz-\lambda(x+y+z-6)$
$L_x=yz-\lambda=0 \Rightarrow \lambda=yz$
$L_y=xz-\lambda=0 \Rightarrow \lambda=xz$
$L_z=xy-\lambda=0 \Rightarrow \lambda=xy$
$yz=xz=xy \Rightarrow x=y=z$
$3x=6 \Rightarrow x=y=z=2$
$\boxed{x=y=z=2}$ noktasında $f_{\text{max}}=8$

Çözüm: İki kısıt için iki çarpan: $L=xyz-\lambda(x+y+z-6)-\mu(x^2+y^2-z^2)$
FOC: $L_x=yz-\lambda-2\mu x=0$
$L_y=xz-\lambda-2\mu y=0$
$L_z=xy-\lambda+2\mu z=0$
$L_\lambda=-(x+y+z-6)=0$, $L_\mu=-(x^2+y^2-z^2)=0$
Çözüm: $x=y=2,\ z=2$ (kontrol: $4+4=8=z^2$ sağlanır). $\boxed{(2,2,2)}$ noktası $f=8$ ile ekstremumdur.

Çözüm: $L=x^2+4y^2-\lambda(x^2+y^2-1)$
$L_x=2x-2\lambda x=2x(1-\lambda)=0$
$L_y=8y-2\lambda y=2y(4-\lambda)=0$
$L_\lambda=-(x^2+y^2-1)=0$
Durum 1: $x=0 \Rightarrow y=\pm1$, $f=4$
Durum 2: $y=0 \Rightarrow x=\pm1$, $f=1$
Durum 3: $x\neq0,\ y\neq0 \Rightarrow \lambda=1$ ve $\lambda=4$ çelişki.
$\boxed{f_{\text{max}}=4}$ (0,±1), $\boxed{f_{\text{min}}=1}$ (±1,0)

Çözüm: $f(x,y)=xy$ (alan), $g(x,y)=2x+2y-20=0$
$L=xy-\lambda(2x+2y-20)$
$L_x=y-2\lambda=0 \Rightarrow y=2\lambda$
$L_y=x-2\lambda=0 \Rightarrow x=2\lambda$
$x=y$, $2x+2y=20 \Rightarrow 4x=20 \Rightarrow x=5$, $y=5$
$\boxed{\text{Maksimum alan }25\text{ birim}^2}$ (kare 5×5)

Çözüm: $f=(x-1)^2+(y-2)^2$ (uzaklığın karesi), $g=x+y-4=0$
$L=(x-1)^2+(y-2)^2-\lambda(x+y-4)$
$L_x=2(x-1)-\lambda=0 \Rightarrow \lambda=2(x-1)$
$L_y=2(y-2)-\lambda=0 \Rightarrow \lambda=2(y-2)$
$x-1=y-2 \Rightarrow x=y-1$
$x+y=4 \Rightarrow (y-1)+y=4 \Rightarrow 2y=5 \Rightarrow y=2.5$, $x=1.5$
$\boxed{(1.5,\ 2.5)}$

Çözüm: $L=x^2+y^2-\lambda(x+y-1)$
$L_{xx}=2$, $L_{xy}=0$, $L_{yy}=2$
$g_x=1$, $g_y=1$
$\tilde{H}=\begin{bmatrix}0&1&1\\1&2&0\\1&0&2\end{bmatrix}$
$\det(\tilde{H})=0-1(2-0)+1(0-2)=-4<0$
$L_{xx}=2>0$ olduğundan $\boxed{\text{minimum}}$

Çözüm: $L=xy-\lambda(x+y-1)$
$L_x=y-\lambda=0$, $L_y=x-\lambda=0$, $L_\lambda=-(x+y-1)=0$
$x=y=0.5$, $\lambda=0.5$
$\lambda=0.5$ demektir: kısıt $x+y=1$'deki 1 birimlik artış ($1\to2$) alanı yaklaşık 0.5 birim artırır. $\boxed{\lambda=0.5}$ (gölge fiyat)

Çözüm: $L=x_1x_2x_3-\lambda(x_1+x_2+x_3-3)$
$L_{x_1}=x_2x_3-\lambda=0$, $L_{x_2}=x_1x_3-\lambda=0$, $L_{x_3}=x_1x_2-\lambda=0$
$x_2x_3=x_1x_3=x_1x_2 \Rightarrow x_1=x_2=x_3$
$3x_1=3 \Rightarrow x_1=x_2=x_3=1$
$\boxed{f_{\text{max}}=1}$

Çözüm: $g(x)=1-x\le0$, $L=x^2+\lambda(1-x)$
KKT: $2x-\lambda=0 \Rightarrow \lambda=2x$
$g(x)\le0 \Rightarrow x\ge1$, $\lambda\ge0$, $\lambda(1-x)=0$
Durum 1: $x=1 \Rightarrow \lambda=2$ (aktif, geçerli)
Durum 2: $\lambda=0 \Rightarrow x=0$ (uygun değil, $x\ge1$ sağlanmaz)
$\boxed{x=1}$ noktasında $f_{\text{min}}=1$

Çözüm: $g(x)=x-1\le0$, $L=x^2+\lambda(x-1)$
KKT: $2x+\lambda=0 \Rightarrow \lambda=-2x$
$g(x)\le0 \Rightarrow x\le1$, $\lambda\ge0$, $\lambda(x-1)=0$
Durum 1: $x=1 \Rightarrow \lambda=-2$ (geçersiz, $\lambda<0$)
Durum 2: $\lambda=0 \Rightarrow x=0$ (uygun, $0\le1$ sağlanır)
$\boxed{x=0}$ noktasında $f_{\text{min}}=0$ (kısıt pasif)

Çözüm: Kenarlar $x,y,z$, yarı çevre $s=6$.
$A^2=s(s-x)(s-y)(s-z)=6(6-x)(6-y)(6-z)$
$L=6(6-x)(6-y)(6-z)-\lambda(x+y+z-12)$
FOC: $(6-y)(6-z)=(6-x)(6-z)=(6-x)(6-y)$ ⇒ $x=y=z$
$3x=12 \Rightarrow x=y=z=4$
$A_{\text{max}}=\sqrt{6\cdot2\cdot2\cdot2}=\sqrt{48}=4\sqrt{3}$
$\boxed{A_{\text{max}}=4\sqrt{3}}$ (eşkenar üçgen)

Çözüm: $f=xyz$, $g=2xy+2xz+2yz-96=0$
$L=xyz-\lambda(2xy+2xz+2yz-96)$
FOC: $yz=2\lambda(y+z)$, $xz=2\lambda(x+z)$, $xy=2\lambda(x+y)$
$x=y=z$ çözümü: $6x^2=96 \Rightarrow x=4$
$\boxed{V_{\text{max}}=4^3=64}$ birim³ (küp)

Çözüm: $L=xy-\lambda(x^2+4y^2-4)$
$L_x=y-2\lambda x=0$, $L_y=x-8\lambda y=0$, $L_\lambda=-(x^2+4y^2-4)=0$
$\lambda=y/(2x)=x/(8y) \Rightarrow 4y^2=x^2 \Rightarrow x=\pm2y$
$x=2y$: $4y^2+4y^2=4 \Rightarrow y=\pm1/\sqrt{2}$, $x=\pm\sqrt{2}$, $f=1$ (maks)
$x=-2y$: $4y^2+4y^2=4 \Rightarrow y=\pm1/\sqrt{2}$, $x=\mp\sqrt{2}$, $f=-1$ (min)
$\boxed{f_{\text{max}}=1,\ f_{\text{min}}=-1}$

Çözüm: $L=K^{0.4}L^{0.6}-\lambda(2K+3L-50)$
FOC: $0.4K^{-0.6}L^{0.6}-2\lambda=0$, $0.6K^{0.4}L^{-0.4}-3\lambda=0$
$\frac{0.4}{2}\frac{L}{K}=\frac{0.6}{3}\frac{K}{L} \Rightarrow 0.2L^2=0.2K^2 \Rightarrow L=K$
$2K+3K=50 \Rightarrow 5K=50 \Rightarrow K=10$, $L=10$
$\boxed{K=10,\ L=10}$ noktasında $Q_{\text{max}}=10$

Çözüm: $L=2x+3y+z-\lambda(x^2+y^2+z^2-14)$
$L_x=2-2\lambda x=0 \Rightarrow x=1/\lambda$
$L_y=3-2\lambda y=0 \Rightarrow y=3/(2\lambda)$
$L_z=1-2\lambda z=0 \Rightarrow z=1/(2\lambda)$
$x^2+y^2+z^2=14 \Rightarrow 1/\lambda^2+9/(4\lambda^2)+1/(4\lambda^2)=14$
$(4+9+1)/(4\lambda^2)=14 \Rightarrow 14/(4\lambda^2)=14 \Rightarrow \lambda^2=1/4 \Rightarrow \lambda=1/2$
$x=2,\ y=3,\ z=1$
$\boxed{f_{\text{max}}=2\cdot2+3\cdot3+1=14}$ (nokta $(2,3,1)$)

Çözüm: $L=x+2y-\lambda(x^2+y^2-25)$
$L_x=1-2\lambda x=0 \Rightarrow x=1/(2\lambda)$
$L_y=2-2\lambda y=0 \Rightarrow y=1/\lambda$
$x^2+y^2=25 \Rightarrow 1/(4\lambda^2)+1/\lambda^2=25 \Rightarrow 5/(4\lambda^2)=25 \Rightarrow \lambda^2=1/20$
$\lambda=1/(2\sqrt5)$: $x=\sqrt5,\ y=2\sqrt5$, $f=\sqrt5+4\sqrt5=5\sqrt5$ (maks)
$\lambda=-1/(2\sqrt5)$: $x=-\sqrt5,\ y=-2\sqrt5$, $f=-5\sqrt5$ (min)
$\boxed{f_{\text{max}}=5\sqrt5,\ f_{\text{min}}=-5\sqrt5}$

Çözüm: $L=x^2+y^2-\lambda(x-y-2)$
$L_x=2x-\lambda=0 \Rightarrow \lambda=2x$
$L_y=2y+\lambda=0 \Rightarrow \lambda=-2y$
$2x=-2y \Rightarrow x=-y$
$x-y=2 \Rightarrow -y-y=2 \Rightarrow y=-1$, $x=1$
$\boxed{(1,-1)}$ noktasında $f_{\text{min}}=2$

Çözüm: $L=x^2+2y^2-\lambda(x+2y-3)$
$L_x=2x-\lambda=0 \Rightarrow \lambda=2x$
$L_y=4y-2\lambda=0 \Rightarrow \lambda=2y$
$2x=2y \Rightarrow x=y$
$x+2y=3 \Rightarrow x+2x=3 \Rightarrow x=1$, $y=1$
$\boxed{(1,1)}$ noktasında $f_{\text{min}}=3$