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Repeat Exercise 1 where $m$ is the minimum value of $f$ on $S$ instead of the maximum value.

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

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 5

Polytopes

Vectors

Harvey Mudd College

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University of Nottingham

Idaho State University

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were asked to repeat exercise one, but now we want to find the minimum value of F on s instead of the maximum value. So recall from exercise one, we had three points P one which was 10 he to which was 23 and p three p three. One second is negative. 12 and said s is the convex hole of P one, p two and P three. And we see that this convex hole is going to be a triangle in the plane. Yeah, and were given linear function ALS and for each functional were asked to find the minimum value now of the functional on this set and then to find all points in the set at which the functional achieves its minimum value. So we see that in part they were given the the near functional F of X one x two equals x one minus X two. So once again we see the F p one is one. The calculations are going to be the same as in exercise. One f of P two is three. I'm sorry. Negative one and f of P three is negative. Three negative three is the smallest of these values so it follows the minimum value is going to be negative three and this occurs only at the point. P three in the set in part B were given the linear functional F of x one x two equals x one plus x two and just as an exercise, one we have the f of p one is equal to one F f p two is five f f p. Three is one. We see that the minimum of thes values has one, and so M is going to be one. And this occurs at points p one p three And in fact, it occurs for all points on the convex hole of P one and p three. Finally, in part C, we're giving the linear functional F of x one x two equals negative three x one plus x two and as an exercise, one we have that f of p one is negative. Three f f p two is negative three and F of P three is equal to five. You see that the minimum of these three numbers is negative. Three. And so it follows that the minimum M is negative three and this will be achieved at all points in the convex hole of P one and P two

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