Chapter 10.9, Problem 38E

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.

  $\{\begin{array}{l}y<x+6\\ 3x+2y\ge 12\\ x-2y\le 2\end{array}$

Answer to Problem 38E

The set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(0,6\right)\\ \left(x_2,y_2\right)=\left(\frac{7}{2},\frac{3}{4}\right)\end{array}$

  

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

Explanation of Solution

Given:

  $\begin{array}{l}y<x+6\\ 3x+2y\ge 12\\ x-2y\le 2\end{array}$

Calculations:

The given eqns. are:

  $y<x+6$.......(1)

  $3x+2y\ge 12$.......(2)

  $x-2y\le 2$.......(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):

  $\begin{array}{l}y<x+6\\ 0<0+6\\ 0<6\end{array}$ ; it satisfy the given equation.

For eq. (3):

  $\begin{array}{l}x-2y\le 2\\ 0-2\left(0\right)\le 2\\ 0\le 2\end{array}$ ;

It also satisfies the given equation.

Clearly, need to shade the region on or below the solid line which have the point $\left(0,0\right)$ .

For eq. (2):

  $\begin{array}{l}3x+2y\ge 12\\ 0+0\ge 12\\ 0\ge 12\end{array}$ ;

It does not satisfy the given equation.

Clearly, 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:

  $y=x+6$.......(4)

  $3x+2y=12$.......(5)

  $x-2y=2$.......(6)

Solve first two eqns:

Put $y=x+6$ in eq. (5):

  $\begin{array}{l}3x+2y=12\\ 3x+2\left(x+6\right)=12\\ 3x+2x+12=12\\ x=0\end{array}$

Now, $y=x+6=0+6=6$

So, the first vertex point is $\left(x_1,y_1\right)=\left(0,6\right)$ .

Solve two eqns (5) and (6):

  $\begin{array}{l}3x+2y+x-2y=2+12\\ 4x=14\\ x=\frac{14}{4}\\ x=\frac{7}{2}\end{array}$

Put $x=\frac{7}{2}$ in eq. (6):

  $\begin{array}{l}x-2y=2\\ \frac{7}{2}-2y=2\\ y=\frac{3}{4}\end{array}$

So, the second vertex point is $\left(x_2,y_2\right)=\left(\frac{7}{2},\frac{3}{4}\right)$ .

So, the set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(0,6\right)\\ \left(x_2,y_2\right)=\left(\frac{7}{2},\frac{3}{4}\right)\end{array}$

  

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.