Exercise 6.2.1

Answers

(a)
$L(x, $λ) = \frac{1}{2} x^2_1 + 2 x^2_2 + λ (x_1 + 3 x_2 - b) $$
(a)
We have
- $\frac{\partial L}{\partial x_{1}} = x_{1} + λ = 0$
- $\frac{\partial L}{\partial x_{2}} = 4 x_{2} + 3 λ = 0$
- $\frac{\partial L}{\partial λ} = x_{1} + 3 x_{2} - b = 0$
(a)
We find $x_{1}^{*} = \frac{4}{13} b, x_{2}^{*} = \frac{3}{13} b, λ^{*} = - \frac{4}{13} b$
(a)
TODO
(a)
The minimum cost is $F^{*} = \frac{2}{13} b^{2}$ , so $\frac{\partial F^{*}}{\partial b} = \frac{4}{13} b = - λ^{*}$

niuers

2020-03-20 00:00