Chapter 10.9, Problem 27E

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.

  $\{\begin{array}{l}x\ge 0\\ y\ge 0\\ 3x+y\le 15\\ 3x+2y\le 9\end{array}$

Answer to Problem 27E

The set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(\frac{5}{3},2\right)\\ \left(x_2,y_2\right)=\left(0,0\right)\\ \left(x_3,y_3\right)=\left(0,3\right)\end{array}$

  

All solutions, or solution set is in the shaded plane so, the solution of the given inequalities is bounded.

Explanation of Solution

Given:

  $\begin{array}{l}x\ge 0\\ y\ge 0\\ 3x+y\le 15\\ 3x+2y\le 9\end{array}$

Calculations:

The given eqns. are:

  $3x+y\le 15$.......(1)

  $3x+2y\le 9$.......(2)

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):

  $\begin{array}{l}3x+y\le 15\\ 0+0\le 15\end{array}$ ; it satisfy the given equation.

For eq. (2):

  $\begin{array}{l}3x+2y\le 9\\ 0+0\le 9\\ 0\le 9\end{array}$ ;

It also satisfies the given equation.

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

From given data points: $x\ge 0,y\ge 0$ :

The values of $x,y$ are non-zero and the solid lines are nothing but the boundary lines. So, the graph of the given inequality equations is:

  

Now, the determine the vertex:

  $x=0$.......(3)

  $y=0$.......(4)

  $3x+5y=15$.......(5)

  $3x+2y=9$.......(6) The first vertex is the intersection point of two lines shown by eq. (5) and (6):

  $\begin{array}{l}3x+5y-3x-2y=15-9\\ 3y=6\\ y=2\end{array}$

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

  $\begin{array}{l}3x+2y=9\\ 3x+2\left(2\right)=9\\ 3x=5\\ x=\frac{5}{3}\end{array}$

So, the first vertex point is $\left(x_1,y_1\right)=\left(\frac{5}{3},2\right)$ .

Now, $x=0,y=0$ so, the second vertex is $\left(x_2,y_2\right)=\left(0,0\right)$ .

The other vertex is the value of the $y$ intercept with the line of equation $3x+5y=15$ :

  $\begin{array}{l}3x+5y=15\\ 3\left(0\right)+5y=15\\ 5y=15\\ y=3\end{array}$

So, $y$ intercept is $\left(0,3\right)$ .

So, the set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(\frac{5}{3},2\right)\\ \left(x_2,y_2\right)=\left(0,0\right)\\ \left(x_3,y_3\right)=\left(0,3\right)\end{array}$

  

All solutions, or solution set is in the shaded plane so, the solution of the given inequalities is bounded.

Conclusion:

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