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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 $\mathbf{x} \geq \mathbf{0}$ . $$A=\left[\begin{array}{ll}{2} & {3} \\ {4} & {1}\end{array}\right], \quad \mathbf{b}=\left[\begin{array}{l}{18} \\ {16}\end{array}\right]$$

$\left\{\left[\begin{array}{l}{0} \\ {0}\end{array}\right],\left[\begin{array}{l}{4} \\ {0}\end{array}\right],\left[\begin{array}{l}{3} \\ {4}\end{array}\right],\left[\begin{array}{l}{0} \\ {6}\end{array}\right]\right\}$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 5

Polytopes

Vectors

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were given a matrix and a column vector were asked to find the minimal representation of the polito to find by the inequalities a X less than equal to be and X greater than or equal to zero. Where is the matrix and B is the column Vector Matrix A Is 2341 column. Vector B is 18 16 and we require that a X is less than or equal to be. And that X is greater than or equal to zero Second requirement really tells us, were restricted to the first quadrant, so mhm we see that the two inequalities are two x one plus three. X two is less than or equal to 18. Call us inequality A and their second inequality be is four x one plus x two is less than or equal to 16. This is what X let's drink would be means. Now we see that the line for inequality A. This goes from the point zero six to the point 90 We see that line be goes from 0 16 to the point 40 now, because X is greater than or equal to zero. We have to include the possibility the X zero and therefore we had that the origin is a vertex. Now the X one intercepts the user. One x two is equal to zero. These are nine and four. So it follows that 40 is a Vertex. Likewise, the X two intercepts which are when x one is equal to zero. These are six and 16 so it follows that 06 is also a vertex. Now let's find where the two lines intersect. We can do this quickly. Looking at her two lines. Take the equation A and then we're going thio multiply it by two and subtracted from equation Be so we get one minus six and negative five x two is equal to 16 and then minus 36 is negative 20 So negative five x two equals negative 20 or X two equals four And therefore, after arranging, we get the X one is three. So I see it. The two lines intersect at the 0.34 so it follows that 34 is also a vertex and therefore the minimal representation for this set is vector 00 Vector 40 Vector 06 and the Vector 34

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