- Consider the constraint $S = \{(x,y) \in \mathbb{R}^2\mid y \geq 0, y \leq 1 - x^2 \}$
Determine the maximum and the minimum of $x + 2y$ when restricted to $S$. Are both at a vertex?
So the way I am interpreting this is I have to use Lagrange Multiple rule, but depending on how to approach this can change the process of how I attack this.
So there is a constraint of $y \leq 1 - x^2 $, but is the constraint $ y \geq 0$ a "constraint".
I know that with one constraint $$\nabla f = \lambda \nabla g$$ but if you have two constraints $$ \nabla f = \lambda \nabla g + \mu \nabla h $$
So with the inequality $y \geq 0$ I know that this is a constraint, but it has no value associated with $x,y,z$, so are all of the partial derivatives just equal to $0$?
You can calculate the gradient of the function for the interior part of the constraint $\{0<y<1-x^2\Rightarrow -1<x<1\}$ and you get $\nabla f(x,y)=(1,2)\ne 0$ so we have not max or min. Now we can investigate the boundaries: $\{y=0,y=1-x^2\Longrightarrow x=\pm1\}$, so we have $(\pm1,0)$; the other boundary $\{y>0,y=1-x^2\Longrightarrow -1<x<1\}$ where the function becomes: $f(x,y=1-x^2)=x+2(1-x^2)=x+2-2x^2$ and putting the derivative equal zero we have $f'(x)=1-4x=0\Longrightarrow x={1\over4}$ that it is a maximum because $f''(x)=-4<0$. So we get $({1\over4},{15\over16})$ as maximum and $(-1,0)$ as minimum, in fact $f(\pm1,0)=\pm1$ and $f\left({1\over4},{15\over16}\right)={17\over8}$.