Exercise 8.10

Answers

\begin{array}{l} \frac{\partial L}{\partial u_{1}} & = 2 u_{1} - α_{1} - α_{2} = 0 \\ \frac{\partial L}{\partial u_{2}} & = 2 u_{2} - 2 α_{1} - α_{3} = 0 \end{array}

Let the derivatives equal to 0, we have $u_{1} = \frac{α_{1} + α_{2}}{2}, u_{2} = \frac{2 α_{1} + α_{3}}{2}$ .

\begin{array}{l} L (u, α) & = u_{1}^{2} + u_{2}^{2} + α_{1} (2 - u_{1} - 2 u_{2}) - α_{2} u_{1} - α_{3} u_{2} \\ = \frac{{(α_{1} + α_{2})}^{2}}{4} + \frac{{(2 α_{1} + α_{3})}^{2}}{4} + α_{1} (2 - \frac{α_{1} + α_{2}}{2} - (2 α_{1} + α_{3})) - α_{2} \frac{α_{1} + α_{2}}{2} - α_{3} \frac{2 α_{1} + α_{3}}{2} \\ = \frac{1}{4} α_{1}^{2} + \frac{1}{2} α_{1} α_{2} + \frac{1}{4} α_{2}^{2} + α_{1}^{2} + α_{1} α_{3} + \frac{1}{4} α_{3}^{2} + 2 α_{1} - \frac{1}{2} α_{1}^{2} \\ - \frac{1}{2} α_{1} α_{2} - 2 α_{1}^{2} - α_{1} α_{3} - \frac{1}{2} α_{1} α_{2} - \frac{1}{2} α_{2}^{2} - α_{1} α_{3} - \frac{1}{2} α_{3}^{2} \\ = - \frac{5}{4} α_{1}^{2} - \frac{1}{4} α_{2}^{2} \frac{1}{4} α_{3}^{2} - \frac{1}{2} α_{1} α_{2} - α_{1} α_{3} + 2 α_{1} \end{array}

niuers

2021-12-08 10:10