Question

Question 1: Given points \({{\mathop{\rm p}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right),{\rm{ }}{{\mathop{\rm p}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right),\) and \({{\mathop{\rm p}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right)\) in \({\mathbb{R}^2}\), let \(S = {\mathop{\rm conv}\nolimits} \left\{ {{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3}} \right\}\). For each linear functional \(f\), find the maximum value \(m\) of \(f\), find the maximum value \(m\) of \(f\) on the set \(S\), and find all points x in \(S\) at which \(f\left( {\mathop{\rm x}\nolimits} \right) = m\).

a.\(f\left( {{x_1},{x_2}} \right) = {x_1} - {x_2}\)

b. \(f\left( {{x_1},{x_2}} \right) = {x_1} + {x_2}\)

c. \(f\left( {{x_1},{x_2}} \right) = - 3{x_1} + {x_2}\)

Short answer

1. \({{\mathop{\rm p}\nolimits} _1}\) is the point in \(S\) at which \(m = 1\).
2. \({{\mathop{\rm p}\nolimits} _2}\)is the point in \(S\) at which \(m = 5\).
3. \({{\mathop{\rm p}\nolimits} _3}\) is the point in \(S\) at which \(m = 5\).

Step-by-step solution

The maximum and minimum is attained at an extreme point

Theorem 16states that considered \(f\) as a linear functional defined on a nonempty compact convex set \(S\).

Then there are extreme points \(\widehat {\mathop{\rm v}\nolimits} \) and \(\widehat {\mathop{\rm w}\nolimits} \) of \(S\) such that \(f\left( {\widehat {\mathop{\rm v}\nolimits} } \right) = \mathop {\max }\limits_{{\mathop{\rm v}\nolimits} \in S} f\left( {\mathop{\rm v}\nolimits} \right)\) and \(f\left( {\widehat {\mathop{\rm w}\nolimits} } \right) = \mathop {\min }\limits_{{\mathop{\rm v}\nolimits} \in S} f\left( {\mathop{\rm v}\nolimits} \right)\).

Determine the maximum value \(m\) of \(f\)

According to theorem 16, the maximum value is attained at one of the extreme points of \(S\).

Evaluate f at the extreme point and select the largest value to find \(m\) as shown below:

1. \({f_1}\left( {{{\mathop{\rm p}\nolimits} _1}} \right) = 1\), \({f_1}\left( {{{\mathop{\rm p}\nolimits} _2}} \right) = - 1\), \({f_1}\left( {{{\mathop{\rm p}\nolimits} _3}} \right) = - 3\), therefore, \({m_1} = 1\). Graph the line \({f_1}\left( {{x_1},{x_2}} \right) = {m_1}\) which means that \({x_1} - {x_2} = 1\), and \({\mathop{\rm x}\nolimits} = {{\mathop{\rm p}\nolimits} _1}\) is the only point in \(S\) at which \({f_1}\left( {\mathop{\rm x}\nolimits} \right) = 1\).

b. \({f_2}\left( {{{\mathop{\rm p}\nolimits} _1}} \right) = 1\), \({f_2}\left( {{{\mathop{\rm p}\nolimits} _2}} \right) = 5\), \({f_2}\left( {{{\mathop{\rm p}\nolimits} _3}} \right) = 1\), therefore, \({m_2} = 5\). Graph the line \({f_2}\left( {{x_1},{x_2}} \right) = {m_2}\) which means that \({x_1} + {x_2} = 5\), and \({\mathop{\rm x}\nolimits} = {{\mathop{\rm p}\nolimits} _2}\) is the only point in \(S\) at which \({f_2}\left( {\mathop{\rm x}\nolimits} \right) = 5\).

c. \({f_3}\left( {{{\mathop{\rm p}\nolimits} _1}} \right) = - 3\), \({f_3}\left( {{{\mathop{\rm p}\nolimits} _2}} \right) = - 3\), \({f_3}\left( {{{\mathop{\rm p}\nolimits} _3}} \right) = 5\), therefore, \({m_3} = 5\). Graph the line \({f_3}\left( {{x_1},{x_2}} \right) = {m_3}\) which means that \( - 3{x_1} + {x_2} = 5\), and \({\mathop{\rm x}\nolimits} = {{\mathop{\rm p}\nolimits} _3}\) is the only point in \(S\) at which \({f_3}\left( {\mathop{\rm x}\nolimits} \right) = 5\).