Question

Question 3: Repeat Exercise 1 where \(m\) is the minimum value of f on \(S\) instead of the maximum value.

Short answer

1. \({{\mathop{\rm p}\nolimits} _3}\) is the point in \(S\) at \(m = - 3\).
2. \(m = 1\)is the minimum value on the set \({\mathop{\rm conv}\nolimits} \left\{ {{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _3}} \right\}\).
3. \(m = - 3\) is the minimum value on the set \({\mathop{\rm conv}\nolimits} \left\{ {{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2}} \right\}\).

Step-by-step solution

Repeat the given as in Exercise 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. \(f\left( {{x_1},{x_2}} \right) = - 3{x_1} + {x_2}\)

The maximum and minimum is attained at an extreme point

Theorem 16states that let \(f\) be a linear functionaldefined 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 minimum value \(m\) of \(f\)

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

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

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

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

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