Find the highest and lowest points in $\mathbb{R}^3$ of the curve of intersection of the plane $2x_1 + 4x_3 = 5$ and the surface $x_1^2 + x_2^2 = 2x_2$.
As far as I understand, the constraint is the intersection : $2x_1 + 4x_3 - 5 = x_1^2 + x_2^2 - 2x_2$ so $g(x_1, x_2) = - x_1^2 - x_2^2 + 2x_1 + 2x_2 + 4x_3 - 5.$
But I don't understand which function should I optimize.
EDIT :
Function to optimize : $f(x_3) = x_3$
Constraint : $\{x \in \mathbb{R}^3 | g(x) = (0, 0)\}$ where $g(x) = (2x_1 + 4x_3 - 5, x_1^2 + x_2^2 - 2x_2)$
We look for $\alpha \in \mathbb{R}$ and $\gamma \in \mathbb{R}^2$ with $\alpha^2 + \|\gamma\|^2 \neq 0$ such that $\alpha Df(a)[h] = (\gamma | Dg(a)[h])$ for each $h \in \mathbb{R}^3$.
a) $h = e_1 = (1, 0 ,0)$
$D(f(x))[e_1] = \frac{\partial{f(x)}}{\partial{x_1}} = 0$
$D(g(x))[e_1] = (\frac{\partial{g_1(x)}}{\partial{x_1}}, \frac{\partial{g_2(x)}}{\partial{x_1}}) = (2, 2x_1)$
b) $h = e_2 = (0, 1 ,0)$\
$D(f(x))[e_2] = \frac{\partial{f(x)}}{\partial{x_2}} = 0$
$D(g(x))[e_2] = (\frac{\partial{g_1(x)}}{\partial{x_2}}, \frac{\partial{g_2(x)}}{\partial{x_2}}) = (0, 2x_2 - 2)$
c) $h = e_3 = (0, 0 ,1)$
$D(f(x))[e_3] = \frac{\partial{f(x)}}{\partial{x_3}} = 1$
$D(g(x))[e_3] = (\frac{\partial{g_1(x)}}{\partial{x_3}}, \frac{\partial{g_2(x)}}{\partial{x_3}}) = (4, 0)$
We obtain the system :
\begin{cases} 0 = (\gamma | (2, 2x_1)) = 2\gamma_1 + 2x_1\gamma_2 \\ 0 = (\gamma | (0, 2x_2 - 2)) = (2x_2 - 2)\gamma_2 \\ \alpha = (\gamma | (4, 0)) = 4\gamma_1 \\ 2x_1 + 4x_3 = 5 \\ x_1^2 + x_2^2 = 2x_2 \\ \end{cases}
But here i'm on lost. How to resolve that system. I would like to do something like let $\lambda = \frac{\alpha}{\gamma}$ but it does not seem to work.
You should optimize $$ \min / \max x_3$$ subject to
$$2x_1+4x_3=5$$
$$x_1^2+x_2^2=2x_2$$
Edit:
$$L(x, \lambda) = x_3 + \lambda_1(2x_1+4x_3-5)+\lambda_2(x_1^2+x_2^2-2x_2)$$
$$2\lambda_1+2\lambda_2x_1=0\tag{1}$$
$$2\lambda_2x_2-2=0\tag{2}$$
$$1+4\lambda_1=0\tag{3}$$
$$3x_1+4x_3=5\tag{4}$$
$$x_1^2+x_2^2=2x_2\tag{5}$$
From equation $(3)$, $\lambda_1=-\frac14$.
$$x_1=\frac1{2\lambda_2}$$
$$x_2=\frac1{\lambda_2}$$
Substitute this into equation $(5)$, we should be able to solve for $\lambda_2$.
After solving for $\lambda_2$, you should be able to recover $x_1$ and $x_2$ and hence $x_3$.