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For any $\rho$ we want to minimize this function. The minimum of $\pi$ is obtained at a point where $x_2 = \frac{1}{2}x_{1}$ and where $x_1$ minimizes the function defined by $$\begin{cases} \frac{1}{4}x_{1}^{2}-\rho(x_{1}-1),& \text{if} \ x_1 <1 \\\\ \frac{1}{4}x_{1}^2,&\text{if} \ x_{1} \geq 1 \end{cases}$$

How do we know that for any $\rho >\frac{1}{2}$, $x_1 = 1$ is a minimum of the function? How did we get the $\frac{1}{2}$? By taking the derivative of those functions?

Added. We want the left branch to decrease if $x_1 <1$ and the right branch to increase for $x_1 \geq 1$. This means that $x_1=1$ is the minimum. It turns out for $\rho > \frac{1}{2}$ this always happens. How do we get the $\frac{1}{2}$?

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1 Answer 1

For your function, $f$, defined by the displayed formula, we have $$ f'(x_1)=\cases{\textstyle{1\over2}x_1 -\rho, &$x_1<1$\cr \vphantom{1\over2}\textstyle{1\over2}x_1, &$x_1>1$ }. $$

If you want your function to be decreasing for $x_1<1$, then the derivative must be negative for $x_1<1$. In order for this to occur, you need $$\textstyle{1\over2}x_1-\rho<0,\quad x_1<1.$$ This condition is satisfied if $\rho>{1\over2}$.

Note that for $x_1>1$, we always have $f'(x_1)>0$, so the right branch is always increasing.

Alternatively, note that you want the $x$ coordinate of the vertex of the parabola ${1\over 4}x^2-\rho(x-1)$ to be greater than or equal to 1. The $x$ coordinate of the vertex is ${-(-\rho)\over 2\cdot{1\over4}}=2\rho$; so, you need $2\rho\ge 1$, or $\rho\ge 1/2$.

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This condition is satisfied for $\rho > \frac{1}{2}$. Why? –  Shawn May 10 '12 at 15:37
vmg got it.......... –  Shawn May 10 '12 at 15:39
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