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In Exercises $5-8,$ find the minimal representation of the polytope defined by the inequalities $A \mathbf{x} \leq \mathbf{b}$ and $\mathbf{x} \geq \mathbf{0}$ . $$A=\left[\begin{array}{ll}{1} & {3} \\ {1} & {1} \\ {4} & {1}\end{array}\right], \mathbf{b}=\left[\begin{array}{l}{18} \\ {10} \\ {28}\end{array}\right]$$
$\left\{\left[\begin{array}{l}{0} \\ {0}\end{array}\right],\left[\begin{array}{l}{7} \\ {0}\end{array}\right],\left[\begin{array}{l}{6} \\ {4}\end{array}\right],\left[\begin{array}{l}{0} \\ {6}\end{array}\right]\right\}$
Calculus 3
Chapter 8
The Geometry of Vector Spaces
Section 5
Polytopes
Vectors
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were given The Matrix and a column vector were asked to find the middle representation of Polito, represented by the Inequalities X is less than or equal to be and X is greater than or equal to zero where A and B are the matrix and column that they're respectively. So The Matrix A is 131141 in the column. Vector B is 18 10 28. So we obtained three inequalities. X one plus three x two is less than or equal to 18. I'll call this inequality a inequality be, which says that X one plus x two is less than or equal to 10 and inequality See which says that four x one plus sex to is less than or equal to 28. Now we see the line A goes from the 0.0 six to the point 18 0 since X greater than equal to zero tells us that Polly Tope is going to be in the first quadrant. We have that mind be goes from 0 10 to the 0.10 0 and finally line See goes from the point c 0 28 to the point 70 now again because X is greater than a quarter zero. We have that 00 is going to be a vertex of the simplex or polito Pennine. Now the X one intercepts these air one x two is zero these air going to be 18 10 and seven. So it follows that 70 is a vertex. Yeah. Now the X two intercepts, which is one X one, is equal to zero. These are 6, 10 and 28. So it follows that 06 is a vertex of the poly tube. Now let's find the intersection of the lines. We see that lines A and B intersect well. We can simply subtract B from A to get two x two equals eight or X two equals four and then from that x one equals six now So we have 64 is a point of intersection and likewise we see that 64 is also a point that lies in line. See, So it follows that all three lines intersect at the 0.6 four so it follows its 64 is a vertex and putting this all together The minimal representation is 00 70 06 and six four
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