Все решила подробно надеюсь поможет
Ответ6п там сложить и все
y = - (x^2+36)/x
y= -(x+36/x)=-x-36/x
y'=-1+36/x^2
-1+36/x^2=0
1-36/x^2=0
(x^2-36)/x^2=0
(x-6)(x+6)/x^2=0
x=-6;x=6;x=0
- + + -
-----(-6)------(0)-----(6)----->x => x=6 - точка максимума
V ^ ^ V
![(3x+y-4)^2+(x+y-2)^2=0](https://tex.z-dn.net/?f=%283x%2By-4%29%5E2%2B%28x%2By-2%29%5E2%3D0)
Так как
![(3x+y-4)^2 \geq 0](https://tex.z-dn.net/?f=%283x%2By-4%29%5E2+%5Cgeq+0)
и
![(x+y-2)^2 \geq 0](https://tex.z-dn.net/?f=%28x%2By-2%29%5E2+%5Cgeq+0)
то уравнение верно только при
![\left \{\begin{array}{I} 3x+y-4=0 \\ x+y-2=0 \end{array}}](https://tex.z-dn.net/?f=%5Cleft+%5C%7B%5Cbegin%7Barray%7D%7BI%7D+3x%2By-4%3D0+%5C%5C+x%2By-2%3D0+%5Cend%7Barray%7D%7D)
Решаем
![\left \{\begin{array}{I} 3x+y-4=0 \\ - \\ x+y-2=0 \end{array}}](https://tex.z-dn.net/?f=%5Cleft+%5C%7B%5Cbegin%7Barray%7D%7BI%7D+3x%2By-4%3D0+%5C%5C+-+%5C%5C+x%2By-2%3D0+%5Cend%7Barray%7D%7D+)
![2x-2=0 \\ x=1 \\ \\ 3+y-4=0 \\ y=1](https://tex.z-dn.net/?f=2x-2%3D0+%5C%5C+x%3D1+%5C%5C+%5C%5C+3%2By-4%3D0+%5C%5C+y%3D1)
Ответ:
![(1;\ 1)](https://tex.z-dn.net/?f=%281%3B%5C+1%29)