Revision history [back]

I missed parts of this questions as well (one being my drawing being backwards :( ). But I found my max and min to be : max - (1.0) and min - (-1,0).

I started the same way as you with the $ x= \frac 1{2\lambda}$ and $ 2y = \lambda 2y$ which I solved out to get $ y = \lambda y$. also with my third equation being $ x^2 + y ^2 = 1$

from here I solved my y equation to find that y = 0. Which looking back now, I'm not sure exactly how I came up with that, so maybe thats where I went wrong as well. But I'll continue..

Using y = 0 I then plugged into our 3rd equation to get $$ \frac 1{2\lambda} ^2 + 0^2 = 1$$ $$ \frac 1{4\lambda ^2} = 1$$ $$ 1 = 4\lambda ^2 $$ $$ \frac 14 = \lambda ^2$$ $$ \sqrt \frac 14 = \lambda$$ $$ \lambda = ^+ _- \frac 12$$

From here I went back and plugged $^+_- \frac 12$ back into our first two equations to find that... $$\frac 1{2(\frac 12)} = x = 1$$ and $$ \frac 1{2(-\frac 12)} = x = -1$$ $$ y = 0$$

I hope that this will help at least a little bit, or if someone could clarify where I went wrong as well.