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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.

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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 $x_3$.

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  • $\begingroup$ What do you mean by "min / max $x_3$" ? $\endgroup$ – Mathieu May 13 at 7:43
  • $\begingroup$ If you want to minimize height, do $\min x_3$. If you want to maximize the height, do $\max x_3$. $\endgroup$ – Siong Thye Goh May 13 at 7:48
  • $\begingroup$ You mean maximize and minimize $f(x_3) = x_3$ under the constraint $g(x) = (2x_1+4x_3-5, x_1^2+x_2$2-2x_2) ?$ $\endgroup$ – Mathieu May 16 at 17:00
  • $\begingroup$ yes, that is correct. $\endgroup$ – Siong Thye Goh May 16 at 17:01
  • $\begingroup$ Thanks for the help ! $\endgroup$ – Mathieu May 16 at 17:02

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