Find extrema of $f(x, y) = x - x^2 - xy^2$ with constraints $\{(x, y) : x^2 + y^2 \le 4, x \ge 0\}$? I know it's a bit short ,
But I have no idea at all how to solve such question ..
I'd be very glad if you can give me some help HOW to solve it.
Given fuction $f(x, y) = x - x^2 - xy^2$, and I need to find it's minimum and maximum value in the domain $D = \{(x, y) : x^2 + y^2 \le 4, x \ge 0\}$.
How can I solve it?
Is that somehow related with Lagrange multipliers?
 A: You do not have to use Lagrange Multipliers, although you should repeat this exercise in order to duplicate the answer using that approach.
We are given:
$$f(x, y) = x - x^2 - xy^2, ~~D = \{(x, y) : x^2 + y^2 \le 4, x \ge 0\}$$
We need to find all of the candidate Critical Points (CPs).


*

*Interior CPs: 


$$f_x = 1-2x -y^2 = 0, f_y = -2xy = 0 \rightarrow (x,y) = \left(\dfrac{1}{2},0\right), (0, 1), (0, -1)$$


*

*Boundary CPs:


$$D = \{(x, y) : x^2 + y^2 \le 4\} \rightarrow y \le \pm~\sqrt{4-x^2}~~\mbox{with endpoints}~~(x,y) = (0, \pm~ 2)$$


*

*$f(x,y)$ becomes $$f = x-x^2-xy^2 = x-x^2-x(4-x^2) = 0 \rightarrow x = \dfrac{1}{3}(1 \pm ~ 10)$$


We reject the negative $x$ point because of the constraint that $x \ge 0$, so $y = \pm~ \sqrt{4-x^2} = \pm ~ 1.4405$.
Thus, we have the seven critical points to test by plugging them in $f(x, y)$. For the four points where $x = 0$, the function value equals zero, so we only need to test the other three points.
We find:


*

*Global Maximum at $(x, y) = \left(\dfrac{1}{2},0\right)$, that is $\dfrac{1}{4}$.

*Global Minimum at $(x, y) = (1.38743, -1.4405)$ and $(1.38743, 1.4405)$, that is $-3.4165$.

A: In the interior of the semicircle, use the usual test, namely, you want $Df(x,y)=0$. In the boundary, you need to maximize or minimize a one variable function: either set $x=0$ or $x^2+y^2=4$ and substitute.
