calculusfunctionsmultivariable-calculusoptimization
I have the function$ f(x,y)=(x+y-2)^2$
I have the constraints $0\leq x \leq 3$ and $x \leq y \leq 3$
The partial derivatives are $f_x =2(x+y-2)$ and $f_y =2(x+y-2)$
So the stationary points in the domain are (0,2) and (1,1)
Now if I look at the second partial derivative it isn't helpful at $f_{xx} =2 f_{yy}=2$ and $f_{xy} =2$
How can I find the absolute maximum and minimum thanks
I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.
edit 1:(add more details)
$f(x,y)=(x-2)^2+y^2+5 \implies (x-2)^2+y^2=f(x,y)-5=R^2$
you want to find max and min of $f(x,y)$ is same to find max and min of $R$.
note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$
if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.
When differentiating implicitly you should get that: \begin{align*} y'(x) = - \frac{\frac{\partial F}{\partial x}(x,y(x))}{\frac{\partial F}{\partial y}(x,y(x))} = \frac{2x-4x^3}{2y-4y^3} \end{align*} Now at an extremum we have that $y'(x)=0$. Lets assume that $2y-4y^3 \neq 0 $ (if we want $y$ to be an implicit function of $x$ then this is a demand). Then $y'(x)=0 \Leftrightarrow 2x-4x^3 =0 \Leftrightarrow 2x(1-2x^2)=0 \Leftrightarrow x =0 \vee x= \pm \sqrt{\frac{1}{2}}$.
Now just check the values of $y$ at these $x$-values (and possible endpoints of the domain of definition of $y$).
Best Answer
You'll have that all of the points $(x,y)$ with $x+y = 2$ will be critical points, not only $(1,1)$ and $(0,2)$. The function is zero at these points and $f\geq 0 $ always, so each one of them is an absolute minimum.
You can also look at $(x+y-2)^2$ with $x+y$ as a variable on its own. Given that $0\leq x\leq y \leq 3$, its greater value when $x=y=3$, so $(3, 3)$ is the maximum point with $f(3,3) = 16 $.