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Pavel [41]
3 years ago
10

The vertices of the feasible region are (choose all that apply)

Mathematics
1 answer:
Anarel [89]3 years ago
5 0

9514 1404 393

Answer:

  (0,0)

  (0,3)

  (3,2)

  (4,0)

Step-by-step explanation:

The feasible region is the doubly-shaded area in the lower left corner of the first quadrant. It is a quadrilateral, so has 4 vertices. Those are the vertices of the feasible region:

  (0,0)

  (0,3)

  (3,2)

  (4,0)

_____

<em>Additional comment</em>

The objective function has positive slope, so the point (4, 0) will maximize it, and the point (0, 3) will minimize it.

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brilliants [131]

Answer:

<em>(a) x=2, y=-1</em>

<em>(b)  x=2, y=2</em>

<em>(c)</em> \displaystyle x=\frac{5}{2}, y=\frac{5}{4}

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.

We call the determinant of the system

\Delta=\begin{vmatrix}a &b \\c  &d \end{vmatrix}

We also define:

\Delta_x=\begin{vmatrix}p &b \\q  &d \end{vmatrix}

And

\Delta_y=\begin{vmatrix}a &p \\c  &q \end{vmatrix}

The solution for x and y is

\displaystyle x=\frac{\Delta_x}{\Delta}

\displaystyle y=\frac{\Delta_y}{\Delta}

(a) The system to solve is

\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.

Calculating:

\Delta=\begin{vmatrix}1 &1 \\1  &-2 \end{vmatrix}=-2-1=-3

\Delta_x=\begin{vmatrix}1 &1 \\4  &-2 \end{vmatrix}=-2-4=-6

\Delta_y=\begin{vmatrix}1 &1 \\1  &4 \end{vmatrix}=4-3=3

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1

The solution is x=2, y=-1

(b) The system to solve is

\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.

Calculating:

\Delta=\begin{vmatrix}4 &-1 \\1  &-1 \end{vmatrix}=-4+1=-3

\Delta_x=\begin{vmatrix}6 &-1 \\0  &-1 \end{vmatrix}=-6-0=-6

\Delta_y=\begin{vmatrix}4 &6 \\1  &0 \end{vmatrix}=0-6=-6

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2

The solution is x=2, y=2

(c) The system to solve is

\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.

Calculating:

\Delta=\begin{vmatrix}-1 &2 \\1  &2 \end{vmatrix}=-2-2=-4

\Delta_x=\begin{vmatrix}0 &2 \\5  &2 \end{vmatrix}=0-10=-10

\Delta_y=\begin{vmatrix}-1 &0 \\1  &5 \end{vmatrix}=-5-0=-5

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}

The solution is

\displaystyle x=\frac{5}{2}, y=\frac{5}{4}

(d) The system to solve is

\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.

Calculating:

\Delta=\begin{vmatrix}6 &-1 \\4  &-2 \end{vmatrix}=-12+4=-8

\Delta_x=\begin{vmatrix}-5 &-1 \\6  &-2 \end{vmatrix}=10+6=16

\Delta_y=\begin{vmatrix}6 &-5 \\4  &6 \end{vmatrix}=36+20=56

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7

The solution is x=-2, y=-7

4 0
3 years ago
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