Question

Решите задачу №3,4 пожалуйста

Answer 1

$1)\; \; M_1(1,1,1)\; \; ,\; \; M_2(5,-4,8)\; \; ,\; \; \vec{l}=\ovecrline {M_1M_2}=(4,-5,7)\\\\U=ln(1+x^2-y^2+z^2)\\\\\frac{\partial U}{\partial x}=\frac{2x}{1+x^2-y^2+z^2} \; \; ,\; \; \frac{\partial U}{\partial x}\Big |_{M_1}=\frac{2}{2}=1\\\\\frac{\partial U}{\partial y}=\frac{-2y}{1+x^2-y^2+x^2}\; \; ,\; \; \frac{\partial U}{\partial y}\Big |_{M_1}=\frac{-2}{2}=-1\\\\\frac{\partial U}{\partial z}=\frac{2z}{1+x^2-y^2+z^2y}\; \; ,\; \; \frac{\partial U}{\partial z}\Big |_{M_1}=\frac{2}{2}=1$

$\overline {grad\, U(M_1)}=\frac{\partial U}{\partial x}\cdot \vec{i}+\frac{\partial U}{\partial y}\cdot \vec{j}+\frac{\partial U}{\partial z}\cdot \vec{k}=\vec{i}-\vec{j}+\vec{k}\\\\\\|\overline {M_1M_2}|=\sqrt{4^2+5^2+7^2}=\sqrt{90}=3\sqrt{10}\\\\cos\alpha =\frac{4}{3\sqrt{10}}\; \; ,\; \; cos\beta =\frac{-5}{3\sqrt{10}}\; \; ,\; \; cos\gamma =\frac{7}{3\sqrt{10}}\\\\\frac{\partial U}{\partial \vec{l}}\Big |_{M_1}=1\cdot \frac{4}{3\sqrt{10}}+(-1)\cdot \frac{-5}{3\sqrt{10}}+1\cdot \frac{7}{3\sqrt{10}}=\frac{13}{3\sqrt{10}}$

$2)\; \; z=xy-2x-y\; \; ,\; \; 0\leq x\leq 3\; \; ,\; \; 0\leq y\leq 4\\\\a)\; \; \left \{ {{z'_{x}=y-2} \atop {z'_{y}=x-1}} \right.\; \; \left \{ {{y-2=0} \atop {x-1=0}} \right.\; \; \left \{ {{y=2} \atop {x=1}} \right.\; \; \to \; \; M_1(1,2)\\\\z(M_1)=z(1,2)=1\cdot 2-2\cdot 1-2=\boxed {-2}\\\\b)\; \; y=0:\; \; z=-2x\; \; ,\; \; z'=-2\ne 0\\\\M_2(0,0)\; ,\; \; z(M_2)=\boxed {0}\\\\M_3(3,0):\; \; z(M_3)=\boxed {-6}\\\\c)\; \; x=0:\; \; z=-y\; \; ,\; \; z'=-1\ne 0\\\\M_4(0,4):\; \; z(M_4)=\boxed{-4}$

$d)\; \; M_3M_4:\; \; \frac{x-3}{-3}=\frac{y}{4}\; \; ,\; \; 4x-12=-3y\; \; ,\; \; y=-\frac{4}{3}\, x+4\\\\z=x\cdot (-\frac{4}{3}\, x+4)-2x-(-\frac{4}{3}\, x+4)=-\frac{4}{3}\, x^2+\frac{10}{3}\, x-4\\\\z'=-\frac{8}{3}\, x+\frac{10}{3}=0\; \; \to \; \; \frac{8}{3}\, x=\frac{10}{3}\; \; ,\; \; x=\frac{5}{4}\\\\y(\frac{5}{4})=-\frac{4}{3}\cdot \frac{5}{4}+4=\frac{7}{3}\\\\M_5(\frac{5}{4}\, ,\, \frac{7}{3}):\; \; z(M_5)=-\frac{4}{3}\cdot  \frac{25}{16}+\frac{10}{3}\cdot \frac{5}{4}-4=-\frac{25}{12}+\frac{50}{12}-\frac{4\cdot 12}{12}=-\frac{23}{12}=\boxed {-1\frac{11}{12}}$

$e)\; \; min\limits _{D}\, z=z(M_3)=z(3,0)=-6\\\\max\limits _{D}\, z=z(M_2)=z(),0)=0$