How many extreme values can you find for $x(1-y)$ for $(x,y)$ on the unit cirle $x^2+y^2=1$?
When using parametrization I find three but with Lagrange multiplier only two. What would be correct? And what would the values be?
For parametrization I used:
$x=\cos(t)$
$y=\sin(t)$
Leading to $2\sin^2(t)-\sin(t)-1=0$ for which I used substitution $u=\sin(t)$ $\Rightarrow$ $2u^2-u-1=0$ giving $u=1$ and $u=-\frac{1}{2}$. Not sure how to proceed?
For Lagrange we find $y=1$ and $y=-\frac{1}{2}$ which inserted in $y^2 = y+x^2$ gives $x=+-\frac{\sqrt3}{2}$. Inserted in $x(1-y)$ gives the two values:
$f_1=\frac{3\sqrt3}{2}$
$f_2=-\frac{3\sqrt3}{2}$
You have set up the pullback $\hat f(t):=x(t)\bigl(1-y(t)\bigr)=\cos t(1-\sin t)$ and now have to determine the $t$-values for which $f'(t)=0$. The latter happens iff $\sin t=u\in\bigl\{1,-{1\over2}\bigr\}$. We therefore have to find out for which values of $t$ modulo $2\pi$ we have $\sin t=1$ or $\sin t=-{1\over2}$. As is well known these values are $$t\in\left\{{\pi\over2},{7\pi\over6},{11\pi\over6}\right\}\qquad({\rm mod}\ 2\pi)$$ (note that $\sin{\pi\over6}=\sin 30^\circ={1\over2}$). We therefore have the three conditionally critical points ${\bf z}_i=(\cos t_i,\sin t_i)\in S^1$ given by $${\bf z}_1=(0,1),\quad{\bf z}_2=\left(-{\sqrt{3}\over2},-{1\over2}\right),\quad{\bf z}_3=\left({\sqrt{3}\over2},-{1\over2}\right)\ .$$ We then can say that $$\max\bigl\{f(x,y)\,\bigm|\,(x,y)\in S^1\bigr\}=\max_{1\leq i\leq3} f({\bf z}_i)\ ,$$ and simlarly for the $\min$. You now have to compute and compare these values.