$z=\mp\sqrt{30-x^2-y^2}$ olur. $z=\sqrt{30-x^2-y^2}$ olsun.
$f(x,y)=x-2y+5\sqrt{30-x^2-y^2}$ olur.
$\dfrac{\partial f}{\partial x}=1-\dfrac{5x}{\sqrt{30-x^2-y^2}}=0\Rightarrow 26x^2+y^2=30$ $(1)$
$\dfrac{\partial f}{\partial y}=-2-\dfrac{5y}{\sqrt{30-x^2-y^2}}=0\Rightarrow 4x^2+29y^2=120$ $(2)$
$(1)$ ve $(2)$ cozulurse
$x=\mp1$ ve $y=\mp2$ elde edilir. $x=\mp1$ ve $y=\mp2$ icin $z=\mp5$ cikar.
Max degeri icin $x=1, y=-2$ ve $z=5$ secilir ve Max degeri olarak $30$ bulunur..
Min degeri icin $x=-1, y=2$ ve $z=-5$ secilir ve Min degeri olarak $-30$ bulunur..
Ikinci turevin isaretine bakilarak da kritik noktalarin lokal max veya lokal min olup olmadigi arastirilabilirdi..
Lagrange carpanlari ile de bulunabilir..
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Lagrange Carpani ile cozelim.
$\mathcal{L}(x,y,z,\lambda)=f(x,y,z)-\lambda g(x,y,z)$
$\mathcal{L}(x,y,z,\lambda)=x-2y+5z-\lambda(x^2+y^2+z^2-30)$
$\begin{array}{lr} \dfrac{\partial\mathcal{L}}{\partial x}=1-2\lambda x=0&&&&&(1 ) \\ \\ \dfrac{\partial\mathcal{L}}{\partial y}=-2-2\lambda y=0 &&&&& (2) \\\\ \dfrac{\partial\mathcal{L}}{\partial z}=5-2\lambda z=0&&&&& (3)\\\\ \dfrac{\partial\mathcal{L}}{\partial \lambda}=x^2+y^2+z^2-30=0&&&&& (4) \end{array}$
$\begin{array}{lr} (1) \implies x=\dfrac{1}{2\lambda}& \\ \\ (2) \implies y=-\dfrac{1}{\lambda}& \\\\ (3) \implies z=\dfrac{5}{2\lambda}& \end{array}$
$x,y$ ve $z$'leri $(4)$'de yerine koyalim.
$\dfrac{1}{4\lambda^2}+\dfrac{1}{\lambda^2}+\dfrac{25}{4\lambda^2}=30\implies \lambda=\mp\dfrac{1}{2}$
$\lambda=\dfrac{1}{2}:\quad (x,y,z)=(1,-2,5)$
$\lambda=-\dfrac{1}{2}:\quad (x,y,z)=(-1,2,-5)$
$f(1,-2,5)=1+2(2)+5(5)=30$ max noktasi.
$f(-1,2,-5)=-1-2(2)+5(-5)=-30$ min noktasi.
