Can Anyone help me with Lagrange multiplier problem

Question

I need to find absolute maximum and minimum of thi function

$$F(x,y) = x^{2} - y^{2} - 2y$$ over $$R = \{ (x,y)\ |\ x^{2} + {y^2} \leq 1\} $$

Thanks for help

Answer 1

$$x^2+y^2 \leq 1 \rightarrow \begin{cases} & \ -1 \leq x \leq1 \\ & \ -1\leq y \leq1 \end{cases}$$now you have f=x^2-y^2-2y $$f=x^2-(y+1)^2+1\\ $$and you have to find max , min f $$x^2+y^2=1 \rightarrow x^2=1-y^2\\f=(1-y^2)-(y+1)^2+1 $$find f' then find roots of f'=0 then put roots in $$ f(y)=(1-y^2)-(y+1)^2+1$$ and chek f (y=-1 , 1) then you will have multiple numbers
greater ids max smaller is min

Answer 2

No Lagrange is useful since the constraint is $x^2+y^2 \le 1.$ [Would need an equality constraint for LaGrange.] You have already found the only critical point of $F(x,y)$ happens to be $(0,-1),$ and this point is coincidentally on the boundary of $x^2+y^2 \le 1.$ The general method, when the region over which the max or min is sought happens to be closed and bounded (as it is here) is to find any critical points inside the region, and also find the max or min on the region's boundary. In your case the boundary may be parametrized for example by $x=\cos t,\ y=\sin t.$ Replace now $x,y$ in your objective function $F(x,y)$ by these formulas, and you have a one variable function to find the max and min of.

When this is done, the critical points come from the factorization of $g'(t)$ where $g(t)=F(\cos(t),\sin(t)).$ You should work it out, but one gets $g'(t)=-2(2\sin t+1)\cos t.$ This has four values of $t$ where $g'(t)=0$ which are fairly simple to express given the factorization.

At the four roots, two give $3/2$, one gives $1$ and the last gives $-3.$ so it seems the max is $3/2$ when $t=-\pi/6,\ 7\pi/6$ and the min is $-3$ when $t=\pi/2.$