Помогите решить, пожалуйста

Question 1

Question 2

Решите задачу:

$z(x,y)=x^2-2xy+\dfrac{5}{2}\, y^2-2x\ \ ,\ \ \ D:\left\{\begin{array}{l}0\leq x\leq 2\\0\leq y\leq 2\end{array}\right\\\\\\1)\ \ \left\{\begin{array}{l}z'_{x}=2x-2y-2=0\\z'_{y}=-2x+5y=0\end{array}\right\ \ \left\{\begin{array}{l}x-y=1\\-2x+5y=0\end{array}\right\ \ \left\{\begin{array}{l}x=y+1\\3y=2\end{array}\right\\\\\\\left\{\begin{array}{l}x=\frac{5}{3}\\\ y=\frac{2}{3}\end{array}\right\ \ \ \ \Rightarrow \ \ \ M_0\Big(\dfrac{5}{3}\, ;\, \dfrac{2}{3}\ \Big)$

$z(M_0)=\dfrac{25}{9}-2\cdot \dfrac{5}{3}\cdot \dfrac{2}{3}+\dfrac{5}{2}\cdot \dfrac{4}{9}-2\cdot \dfrac{5}{3}=\dfrac{25-20+10-20}{9}=\boxed {\, -\dfrac{5}{9}\, }\\\\\\2)\ \ y=0\ ,\ \ x\in [\, 0;2\, ]:\ \ z(x)=x^2-2x\ ;\ \ z'_{x}=2x-2=0\ ,\ \ x=1\\\\M_1(1;0)\ \ ,\ \ z(M_1)=z(1;0)=1-2=\boxed {-1\, }\\\\M_2(0;0)\ \ ,\ \ z(M_2)=z(0;0)={\, \boxed 0\, }\\\\M_3(2;0)\ \ ,\ \ z(M_3)=z(2;0)=4-4=\boxed {\, 0\, }$

$3)\ \ x=2\ ,\ \ y\in [\ 0;2\, ]:\ \ z(y)=4-4y+\dfrac{5}{2}\, y^2-4=\dfrac{5}{2}\, y^2-4y\ ;\\\\z'_{y}=5y-4=0\ \ ,\ \ y=\dfrac{4}{5}\ ,\\\\M_4(2;\dfrac{4}{5})\ \ ,\ \ z(M_4)=z(2;\dfrac{4}{5})=\dfrac{5}{2}\cdot \dfrac{16}{25}-4\cdot \dfrac{4}{5}=-\dfrac{40}{25}=-\dfrac{8}{5}=\boxed {\, -1,6\, }$ $M_5(2;2)\ \ ,\ \ z(M_5)=\dfarc{5}{2}\cdot 4-8=\boxed {\, 2\, }\\\\M_3(2;0)\ \ ,\ \ z(M_3)=z(2;0)=0-0=\boxed {\, 0\, }\\\\\\4)\ \ y=2\ ,\ \ x\in \, [\, 0\, ;2\, ]:\ \ z(x)=x^2-4x+10-2x=x^2-6x+10\ ,\\\\z'_{x}=2x-6=0\ \ ,\ \ x=3\notin [\, 0\, ;\, 2\, ]\\\\M_5(2;2)\ \ ,\ \ z(M_5)=z(2;2)=4-12+10=\boxed{\, 2\, }$

$M_6(0;2)\ \ ,\ \ z(M_6)=z(0;2)=\dfrac{5}{2}\cdot 4=\boxed {\, 10\, }\\\\\\5)\ \ x=0\ ,\ \ y\in [\, 0\, ;\, 2\, ]:\ \ z=\dfrac{5}{2}\, y^2\ \ ,\ \ z'_{x}=5y=0\ \ ,\ \ y=0\\\\M_2(0\, ;\, 0)\ \ ,\ \ z(M_2)=\boxed {\, 0\, }\\\\M_6(0;2)\ \ ,\ \ z(M_6)=z(0;2)=\dfrac{5}{2}\cdot 4=\boxed {\, 10\, }$

$6)\ \ z_{naibol.}\Big|_{D}=z(M_6)=z(0;2)=10\\\\z_{naimen.}\Big|_{D}=z(M_4)=z(2;\dfrac{4}{5})=-1,6$