Solve using Lagrangian method:
$ Maximize \ \ \ (x^2+2y^2-x) \ $ with respect to $ \ x^2+y^2 \leq 1 \ $
Find the $ \ 5 \ $ pairs $ \ (x,y) \ $ that satisfies Lagrangian equations.
Answer: Let $ \ f(x,y)=x^2+2y^2-x \ $ and $ \ g(x,y)=x^2+y^2-1=0 \ $
Let $ \ \lambda \ $ be the lagrangian multiplier.
Then,
$ \nabla f(x)=\lambda \nabla g(x) \\ \Rightarrow \\ 2x-1=2 \lambda x \ \\ 4y=2 \lambda y $
For $ \ y=0 \ $ the second equation satisfies.
Then, $ g(x,0)=0 \Rightarrow x^2=1 \ \Rightarrow x=\pm \ $
Thus the two critical points are $ \ (\pm1,0) \ $
Next, we have to find rest $ \ 3 \ $ critical points.
From the first equation , we have
$ 2x-1=2 \lambda x \ $
But how to find other $ \ 3 \ $ critical points in order to get all $ \ 5 \ $ critical points.
Help me
Presumably the question is to find all possible points that satisfy the necessary first order conditions of optimality.
There are two cases to consider:
(i) $x^2+y^2 <1$, in which case we have $2x-1 = 0, y=0$ which yields $({1 \over 2},0)$.
(ii) $x^2+y^2 =1$. Then the Lagrange conditions give $2x-1=2 \lambda x$, $2y= \lambda y$, or $2x(1-\lambda) = 1$, $(2-\lambda)y = 0$.
Case (a) $\lambda = 2$: Then $x = -{1 \over 2}$ and $y= \pm {\sqrt{3} \over 2}$.
Case(b) $\lambda \neq 2$: Then $y=0$ and $x \pm 1$.
Counting all of these gives five solutions.
Note that since we are maximising, we must have $x\le 0$, so there are really only three solutions to consider, and a quick evaluation shows that the maximisers are $(-{1 \over 2},\pm{\sqrt{3} \over 2})$.