Exercise 1.10.2

Answers

1.
Let $x = (a, b)$ and $y = (c, d)$ , we have $R^{∗} (x) = \frac{5 a^{2} + 8 ab + 5 b^{2}}{a^{2} + 4 b^{2}}$ and $R (y) = \frac{20 c^{2} + 16 cd + 5 d^{2}}{4 (c^{2} + d^{2})}$
1.
Take the $c$ and $d$ derivatives of $R (y)$ to find its maximum and minimum, we have

\begin{array}{l} \frac{∂R (y)}{∂c} & = \frac{4 (40 c + 16 d) (c^{2} + d^{2}) - 8 c (20 c^{2} + 16 cd + 5 d^{2})}{16 (c^{2} + d^{2})} \end{array}

\begin{array}{l} \frac{∂R (y)}{∂d} & = \frac{4 (16 c + 10 d) (c^{2} + d^{2}) - 8 d (20 c^{2} + 16 cd + 5 d^{2})}{16 (c^{2} + d^{2})} \end{array}

Let each derivative equal to 0, we have two equations:

\begin{array}{l} d (8 d^{2} - 8 c^{2} + 15 cd) & = 0 \\ c (8 c^{2} - 8 d^{2} - 15 cd) & = 0 \end{array}

Let $d = 1$ and solve for $c$ have see that the two roots are the same as $y_{1}$ and $y_{2}$ .

1.
Take the $a$ and $b$ derivatives of $R^{∗} (y)$ to find its maximum and minimum, we have

\begin{array}{l} \frac{\partial R^{∗} (y)}{∂a} & = \frac{(10 a + 8 b) (a^{2} + 4 b^{2}) - 2 a (5 a^{2} + 8 ab + 5 b^{2})}{a^{2} + 4 b^{2}} \end{array}

\begin{array}{l} \frac{\partial R^{∗} (y)}{∂b} & = \frac{(10 b + 8 a) (a^{2} + 4 b^{2}) - 8 b (5 a^{2} + 8 ab + 5 b^{2})}{a^{2} + 4 b^{2}} \end{array}

Let each derivative equal to 0, we have two equations:

\begin{array}{l} b (32 b^{2} + 30 ab - 8 a^{2}) & = 0 \\ a (- 32 b^{2} - 30 ab + 8 a^{2}) & = 0 \end{array}

Let $b = 1$ and solve for $a$ have see that the two roots are the same as $x_{1}$ and $x_{2}$ .

niuers

2020-03-20 00:00