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Chapter 8: Q1E (page 437)

Question: 1. Let Lbe the line in \({\mathbb{R}^{\bf{2}}}\) through the points \(\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{4}}\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\end{array}} \right)\). Find a linear functional f and a real number d such that \(L = \left( {f:d} \right)\).

Short Answer

Expert verified

The linear functional is \(f\left( {x,y} \right) = 3x + 4y\), and a real number \(d = 13\) such that \(L = \left( {f:d} \right)\).

Step by step solution

01

Find the line equation

The given two points are \(\left( {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}3\\1\end{array}} \right)\). Take \({x_1} = - 1,{\rm{ }}{y_1} = 4\), and \({x_2} = 3,{\rm{ }}{y_2} = 1\).

The line equation is given by,

\(\begin{array}{c}\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}\\\frac{{y - 4}}{{1 - 4}} = \frac{{x + 1}}{{3 + 1}}\\4y - 16 = - 3x - 3\\3x + 4y = 13\end{array}\)

02

Find the linear functional f

From the above equation, take the linear functional as \(f\left( {x,y} \right) = 3x + 4y\).

03

Conclusion

The line Lcan be written as shown below:

\(\begin{array}{c}L = \left\{ {x \in {\mathbb{R}^2}:3x + 4y = 13} \right\}\\ = \left\{ {x \in {\mathbb{R}^2}:f\left( {x,y} \right) = 13} \right\}\\L = \left( {f:13} \right)\end{array}\)

Thus, the linear function is \(f\left( {x,y} \right) = 3x + 4y\), and a real number \(d = 13\) such that \(L = \left( {f:d} \right)\).

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Most popular questions from this chapter

Question: Repeat Exercise 7 when

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{3}}\\{ - {\bf{2}}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{6}}\\{ - {\bf{5}}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{0}}\\{{\bf{12}}}\\{ - {\bf{6}}}\end{array}} \right)\)

\({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{1}}}\\{{\bf{15}}}\\{ - {\bf{7}}}\end{array}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{5}}}\\{\bf{3}}\\{ - {\bf{8}}}\\{\bf{6}}\end{array}} \right)\), and \({{\bf{p}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{6}}\\{ - {\bf{6}}}\\{ - {\bf{8}}}\end{array}} \right)\)

Question: In Exercises 15-20, write a formula for a linear functional f and specify a number d so that \(\left( {f:d} \right)\) the hyperplane H described in the exercise.

Let A be the \({\bf{1}} \times {\bf{4}}\) matrix \(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{3}}}&{\bf{4}}&{ - {\bf{2}}}\end{array}} \right)\) and let \(b = {\bf{5}}\). Let \(H = \left\{ {{\bf{x}}\,\,{\rm{in}}\,{\mathbb{R}^{\bf{4}}}:A{\bf{x}} = {\bf{b}}} \right\}\).

Question: In Exercises 5-8, find the minimal representation of the polytope defined by the inequalities \(A{\mathop{\rm x}\nolimits} \le {\mathop{\rm b}\nolimits} \) and \({\mathop{\rm x}\nolimits} \ge 0\).

6. \(A = \left( {\begin{array}{*{20}{c}}2&3\\4&1\end{array}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{array}{*{20}{c}}{18}\\{16}\end{array}} \right)\)

Question: In Exercise 3, determine whether each set is open or closed or neither open nor closed.

3. a. \(\left\{ {\left( {x,y} \right):y > {\bf{0}}} \right\}\)

b. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} \le y \le {\bf{3}}} \right\}\)

c. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} < y < {\bf{3}}} \right\}\)

d. \(\left\{ {\left( {x,y} \right):xy = {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

e. \(\left\{ {\left( {x,y} \right):xy \ge {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

14.a. Justify each equal sign:

\(3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\)

b. Show that \({{\mathop{\rm r}\nolimits} _2}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _2}\) to \({{\mathop{\rm p}\nolimits} _3}\).

c. Justify each equal sign: \(3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right)\).

d. Use part (c) to show that \(8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}\).

e. Use part (d) equation (8), and part (a) to show that \({{\mathop{\rm r}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm r}\nolimits} _2}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\). That is, \({{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).
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