given-points-mathbfp_1leftbeginarrayr0-1endarrayright-mathbfp_2leftbeginarrayl2-1endarrayright-and-m

Question

Given points $\mathbf{p}_{1}=\left[\begin{array}{r}{0} \ {-1}\end{array}ight], \mathbf{p}_{2}=\left[\begin{array}{l}{2} \ {1}\end{array}ight],$ and $\mathbf{p}_{3}=\left[\begin{array}{l}{1} \ {2}\end{array}ight]$ in $\mathbb{R}^{2}$ , let $S=\operatorname{conv}\left\{\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}ight\} .$ 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 $f(\mathbf{x})=m .$  
$$
\begin{array}{lll}{	ext { a. }} & {f\left(x_{1}, x_{2}ight)=x_{1}+x_{2}} & {	ext { b. } f\left(x_{1}, x_{2}ight)=x_{1}-x_{2}} \ {	ext { c. }} & {f\left(x_{1}, x_{2}ight)=-2 x_{1}+x_{2}}\end{array}
$$

James Chok · Accepted Answer

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&#x27;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&#x27;re zero last one, if that&#x27;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

Chris Trentman · Answer

we were given three points in our to and were given a poly toke form. From these three points in fact, is the convex closure of these three points her complex whole, and were also given linear function ALS and for each letter functional rest to find the maximum value of the linear functional on the set s and defines all the points and s at which linear functional takes its maximum value. So were given the points P one which is the vector 10 p two, which is the Vector 23 and P three, which is the vector negative 12 So you see the S, which is the convicts hole of P one, p two and P three. This is going to be a triangle in the plane now by a the&#x27;re, um from the book we had that the maximum value of the linear functional f will always occur at one of the extreme points. So in general, for each step, we&#x27;re going to first evaluate the linear functional at each of the extreme points and then select the extreme points that gives you maximum value. So in part, a were given the linear functional F of x one x two equals x one minus X two So we have that. The extreme points of our set s these air the same as the Vergis is of the set s which, because that s is a triangle is the same as points p one p two and p three. So we have that f of p one. This is going to be x one minus x two, which is one f of P two. This is going to be tu minus three, which is negative one and f of P three. This is negative one minus two, which is negative. Three. So it follows that the maximum Value M is equal toe one at point p one. This is the only point at which the functional takes its maximum in part B were given the linear functional F F X one x two equals x one plus x two. So we see that f of p one is going to be one plus zero, which is one F f P two is going to be two plus three, which is five and f of P three is negative one plus two, which is positive one so it follows the maximum value. Am the f takes, It&#x27;s going to be five. It would only take this maximum value at the point p two Now in part C, we&#x27;re given the linear functional F of x one x two equals negative three x one plus x two We see that f of p one. This is going to be negative three plus zero, which is simply negative. Three f of p. Two. This is negative three times to which is negative. Six plus three, which is negative. Three and f of P three is negative three times negative one, which is three plus two, which is five. And so it&#x27;s clear that the maximum m is five and only at the point p three.

Pravakar Paul · Answer

this problem appeared in Partner Competition 2006. So here is the problem. Suppose if it&#x27;s a continuous function from Syria. Cummerbund, close, interfere, toe are and we associate with if to quantity are your fifth, it&#x27;s a call to integration from 0 to 1 exist squared time safe, all fixed E X. Then she&#x27;ll Fif physical to integration from cedar to one six times If I&#x27;ll fix Square DX and I have to find the maximum value off if minors she off if, where if can be any country? What&#x27;s function in zero comma? One close interval. So let&#x27;s compute are your fifth minus G l fifth, which can be bitterness. Integration from 0 to 1 exist square before fix, minus integration from 0 to 1 extents. If our fixed square DX I can combine this toe different integral into a single too integral as integration from 0 to 1 exists quietly for fix minus extents. If affects Claire and followed by t x. No, let&#x27;s try to analyse dysfunction. I have exists kwehr fo fix minus extents. If our fixed square I can rewrite this expression as follows, this is nothing but X times extents. If our fix minus before fixed square. No, it looks like a square, so let&#x27;s complete the square. Let&#x27;s take the negative X out. It becomes a full fix square, minus extents for fix. Now, if I complete this Claire, it can be read Innis Negative IX that I have if x square minus two times a year for fix times X, divided by two plus eggs over two square. And then I can subtract this extra time negative exists Claire, divided by four. So then the entire expression becomes positive. X cubed, divided by four minus X times efore Fix minus X over toe. Hold square now, if affects minus except for two whole skirt. It&#x27;s always say positive number and X is between zero and one. So this in place this inter quantity, it&#x27;s less than a call to X Cube, divided by four, because six times before fix finance. 60 for two Whole square. It&#x27;s always bickered. Chemical co CEO No. Bye monotone. The city off integration we happen are your fifth minus zero Fif. It&#x27;s equality. Integration from 0 to 1 exists Claire times a four fix minus extensive If x square PX. So this is less than a call to indication from 0 to 1 x cube divided by four DX. Now the right hand side becomes one over four times extra. LIBOR four Divided by four. The limit is from 0 to 1, which is his equal to one over 16. So the maximum value off are your fifth minus. She&#x27;ll fif this one over 16 and you just achieved Vin A for fixes a call toe except for two. Salute school back If he for fixes a cold drinks overto, then the same tire expression IFOR fix minus X over to hold squared, it becomes zero. So then it becomes six cubed, divided by four. That&#x27;s the only surviving term. So you end up getting 11 over 16 as my maximum value.

Brandon Fox · Answer

So for this problem, we want to take the F of acts with just two acts of minus 11 plus three times X plus two, and it needs to be, at most zero. So asked me, less than or equal to zero cannot be over zero. So we&#x27;ll saw will distribute this three in here. Let&#x27;s go and distribute the three in so three times X is a plus three x and three times. That, too, is a plus. Six was combined. The light terms two x and three x makes five acts, and the negative 11 plus six makes a minus five. So let&#x27;s get the minus five out of here by adding five double sides and we get five. X is less than or equal to five, and I can divide by five on both sides to get my answer. X is less than or equal to one. So what that means is that my, my, my solution says from negative infinity all the way up into including one. I can put a little bracket because it includes one as well. This is my answer

Ankit Gupta · Answer

Hello everyone. This prison here to find out that&#x27;s what we knew already before doesn&#x27;t even function. So what time to be played? A lot of angry from these two Come on picking minus one bedroom apartment. So if I think that&#x27;s what I for one, didn&#x27;t get it with a square and with me, unless you it&#x27;s and less today it&#x27;s simple. So these are new to get do again. Dignity. What? The brain You have been 100 minus one photo spread minus one. So very personal financial support. Find this 1 may do your X squared plus wait months. One an early. And when I got pregnant, Leslie So what do I get? Minus one by do explicit on this one. Minus on migrant minus one That is less one by do lessening. So I pulled back support while finally home is on this one. But I do it plus one whole square. There it is, So one by you. So I&#x27;ve been getting medical facilities like it also ended maximum radio. So my view is, but this part of it, you know, So it&#x27;s one body. So what is the maximum value and give

Ramzi Deek · Answer

F of X is executed minus two X squared minus 11 X plus 12. So let&#x27;s start off first with finding the zeros off. This probably normal. So that&#x27;s considered 12 here at the end. So what are the factors of 12? One, two, 346 and 12 Right? So what I&#x27;m doing here is called synthetic substitution for synthetic substitution that allows me to find out what the zeros of this function are. So in synthetic substitution, I write down the coefficients first. So I have one negative too negative 11 and 12 as my coefficients. So let&#x27;s take first Ah, number one because it&#x27;s one of the factors of 12 and start doing the synthetic substitution here. So one goes down is 11 times one is one That&#x27;s negative. 11 times negative One is negative. One so minus 12. One times negative 12 is negative 12 zero. And there you go. One is a root of the function. Now what happens here? This becomes X minus one. Since one is a factor or a route X minus, one is a factor and this becomes X squared minus X minus 12. We have two more routes and how can we do that? So let&#x27;s go. Move to the next page here. X minus one into X squared minus X minus 12. If we look here, X squared minus X minus 12 can also be factored. So two numbers that give product of negative 12 and some of negative one. So minus four and positive three. Let&#x27;s see. Minus four plus three is negative one. Okay, that&#x27;s the coefficient of X here. Minus four times three. That&#x27;s negative. 12. So there you go. We have X minus four into X plus three. Now I have the polynomial completely factored. So what are my roots? X is equal to one X is equal to four and X equals two minus three. Those three values are the zeros. Those are what make my function into a zero. Finally moving on to the relative minimum and the relative maximum so relative minimum here and relative maximum. So in this case, you&#x27;re gonna need to graph it on a calculator and using that calculator, it will give you the values off your minimum and maximum. Your minimum should be minus 12.6. Your maximum should be 20 point seven

Problem

Repeat Exercise 1 where $m$ is the minimum value …

Problem 2 Easy Difficulty

Answer

(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 $P_{3} .$

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Linear Algebra and Its Applications

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(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 $P_{3} .$

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Michael J.

(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 $P_{3} .$

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications