Chapter 10.9, Problem 28E

To determine

The given system of equation has to be solved and graphed for its solutions, co-ordinates of all vertices and tells that the set of solution is bounded or not.

  $\{\begin{array}{l}x>2\\ y<12\\ 2x-4y>8\end{array}$

Answer to Problem 28E

The set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(2,-1\right)\\ \left(x_2,y_2\right)=\left(0,0\right)\\ \left(x_3,y_3\right)=\left(4,0\right)\end{array}$

  

All solutions, or solution set is not in the shaded plane so, the solution of the given inequalities is unbounded.

Explanation of Solution

Given:

  $\begin{array}{l}x>2\\ y<12\\ 2x-4y>8\end{array}$

Calculations:

The given eqns. are:

  $x>2$.......(1)

  $y<12$.......(2)

  $2x-4y>8$.......(3)

Use the test point $\left(0,0\right)$ to determine whether which region satisfies the inequalities equation:

Substitute $\left(0,0\right)$ in eq. (1) and (2):

For eq. (1):

  $x>2,0\cancel{>}2$ ; it does not satisfy the given equation. Need to shade the region on or below the solid line which does not have the point $\left(0,0\right)$ .

For eq. (2):

  $\begin{array}{l}y<12\\ 0<12\end{array}$ ;

It also satisfies the given equation. Clearly, need to shade the region on or below the solid line which does have the point $\left(0,0\right)$ .

For eq. (3):

  $\begin{array}{l}2x-4y>8\\ 0-0>8\\ 0\cancel{>}8\end{array}$ ;

It does not satisfy the given equation. Need to shade the region on or below the solid line which does not have the point $\left(0,0\right)$ .

So, the graph of the given inequality equations is:

  

Now, the determine the vertex:

  $x=2$.......(4)

  $y=12$.......(5)

  $2x-4y=8$.......(6) The first vertex is the intersection point of two lines shown by eq. (4) and (6):

Put $x=2$ in eq. (6):

  $\begin{array}{l}2x-4y=8\\ 2\left(2\right)-4y=8\\ -4y=4\\ y=-1\end{array}$

So, the first vertex point is $\left(x_1,y_1\right)=\left(2,-1\right)$ .

The second vertex is the intersection point of two lines shown by eq. (5) and (6):

Put $y=12$ in eq. (6):

  $\begin{array}{l}2x-4y=8\\ 2x-4\left(12\right)=8\\ 2x=56\\ x=28\end{array}$

So, the second vertex is $\left(x_2,y_2\right)=\left(0,0\right)$ .

The other vertex is the value of the $x$ intercept with the line of equation $2x-4y=8$ :

  $\begin{array}{l}2x-4y=8\\ 2x-4\left(0\right)=8\\ 2x=8\\ y=4\end{array}$

So, $x$ intercept is $\left(4,0\right)$ .

So, the set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(2,-1\right)\\ \left(x_2,y_2\right)=\left(0,0\right)\\ \left(x_3,y_3\right)=\left(4,0\right)\end{array}$

  

All solutions, or solution set is not in the shaded plane so, the solution of the given inequalities is unbounded.

Conclusion:

Hence, the given set of equations is derived as above.