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Given points $\mathbf{p}_{1}=\left[\begin{array}{r}{0} \\ {-1}\end{array}\right], \mathbf{p}_{2}=\left[\begin{array}{l}{2} \\ {1}\end{array}\right],$ and $\mathbf{p}_{3}=\left[\begin{array}{l}{1} \\ {2}\end{array}\right]$ in $\mathbb{R}^{2}$ , let $S=\operatorname{conv}\left\{\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}\right\} .$ For each linear functional $f,$ find the maximum value $m$ of $f$ on the set $S,$ and find all points $\mathbf{x}$ in $S$ at which $f(\mathbf{x})=m .$ $$\begin{array}{lll}{\text { a. }} & {f\left(x_{1}, x_{2}\right)=x_{1}+x_{2}} & {\text { b. } f\left(x_{1}, x_{2}\right)=x_{1}-x_{2}} \\ {\text { c. }} & {f\left(x_{1}, x_{2}\right)=-2 x_{1}+x_{2}}\end{array}$$

(a) $\mathrm{m}=3,$ Maximum value is $\mathrm{m}=3$ at point $P_{2}, P_{3}$(b) $\mathrm{m}=1,$ Maximum value is $\mathrm{m}=1$ at point $P_{1}, P_{2}$(c) $\mathrm{m}=0,$ Maximum value is $\mathrm{m}=0$ at point $P_{3} .$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 5

Polytopes

Vectors

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okay in this question, over need to do is on the maximum value F at P. One p to p three. So if you value weighed at P one is equal to zero plus minus, one gives you a modest one if you bother what it at peak, too, is equal to two plus one. Gives you three. F evaluated at p three is equal to one prostitute, which gives you three. So the maximum values is three given at the points he won and p two. For this question, we go f at P one. Is it 20 minus minus? One gives you one. If that's pee, too, is equal to minus one. Gives you one. And if at p three, if you could choose one minus two, which is modest one seven, the maximum value is given by is one at the points p one and P two off with last one F at P one is equal to monetary times. You're zero last one, if that's pee, too. Music minus two times like two miles. Four plus one miles. Three F at P three is equal to minus two times one gives you minds. Two plus two gives you zero, so the maximum value is zero given at points a three

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