Q40. - Algebra 2

Question

Graph each system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the given function for this region.

40. $\begin{array}{c}\mathbf{y}\ge \mathbf{3}\\ \mathbf{y}\le \mathbf{x}\mathbf{+}\mathbf{2}\\ \mathbf{y}\le \mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{15}\\ \mathbf{f}\left(x,y\right)\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\end{array}$

Step-by-Step Solution

Answer

The graph of the given system of inequalities is:

The coordinates of the vertices of the feasible region are: $\mathbf{A}\left(\mathbf{1}\mathbf{,}\mathbf{3}\right)$, $\mathbf{B}\left(\mathbf{6}\mathbf{,}\mathbf{3}\right)$ and $\mathbf{C}\left(\mathbf{4}\mathbf{.}\mathbf{333}\mathbf{,}\mathbf{6}\mathbf{.}\mathbf{333}\right)$.

The maximum and minimum values of the given for the feasible region are $\mathbf{27}\mathbf{.}\mathbf{665}$ and 11 respectively.

Step 1 ­- Description of step.

The first inequality is: $y\ge 3$.

First to draw the inequality, convert inequality into equality.

Therefore, it is obtained that: $y=3$.

Now as it can be seen that $y=3$ is a n equation of a line parallel to $x$-axis and passing through the point $\left(0,3\right)$.

Now as the inequality has equal to sign, therefore the line will be drawn as solid line as it will be also part of the inequality.

Now to draw the inequality $y\ge 3$, take any point which is not on the line $y=3$ and substitute that point in the inequality, if the condition is satisfied then shade the region towards that point otherwise away from the point.

Let the point be $\left(0,0\right)$.

Now substitute that point in the inequality $y\ge 3$.

Therefore, it can be obtained that:

$\begin{array}{c}y\ge 3\\ 0\ge 3\end{array}$

As, the condition obtained is false, therefore to draw the inequality $y\ge 3$, shade the region away from the point $\left(0,0\right)$.

Therefore, the graph of the inequality $y\ge 3$ is:

Step 2 ­- Description of step.

The second inequality is: $y\le x+2$.

First to draw the inequality, convert inequality into equality.

Therefore, it is obtained that: $y=x+2$.

When $x=0$,

 $\begin{array}{c}y=0+2\\ =2\end{array}$

Therefore, the first point is $\left(0,2\right)$

When $x=-2$,

 $\begin{array}{c}y=-2+2\\ =0\end{array}$

Therefore, the second point is $\left(-2,0\right)$

Now it can be seen that $y=x+2$ is the equation of a line passing through the points $\left(0,2\right)$ and $\left(-2,0\right)$.

Now as the inequality has equal to sign, therefore the line will be drawn as solid line as it will be also part of the inequality.

Now to draw the inequality $y\le x+2$, take any point which is not on the line $y=x+2$ and substitute that point in the inequality, if the condition is satisfied then shade the region towards that point otherwise away from the point.

Let the point be $\left(0,0\right)$.

Now substitute that point in the inequality $y\le x+2$.

Therefore, it can be obtained that:

 $\begin{array}{c}y\le x+2\\ 0\le 0+2\\ 0\le 2\end{array}$

As, the condition obtained is true, therefore to draw the inequality $y\le x+2$, shade the region towards the point $\left(0,0\right)$.

Therefore, the graph of the inequality $y\le x+2$ is:

Step 3 ­- Description of step.

The third inequality is: $y\le -2x+15$.

First to draw the inequality, convert inequality into equality.

Therefore, it is obtained that: $y=-2x+15$.

When $x=7$,

 $\begin{array}{c}y=-2\left(7\right)+15\\ =-14+15\\ =1\end{array}$

Therefore, the first point is $\left(7,1\right)$

When $x=8$,

 $\begin{array}{c}y=-2\left(8\right)+15\\ =-16+15\\ =-1\end{array}$

Therefore, the second point is $\left(8,-1\right)$

Now it can be seen that $y=-2x+15$ is the equation of a line passing through the points $\left(7,1\right)$ and $\left(8,-1\right)$.

Now as the inequality has equal to sign, therefore the line will be drawn as solid line as it will be also part of the inequality.

Now to draw the inequality $y\le -2x+15$, take any point which is not on the line $y=-2x+15$ and substitute that point in the inequality, if the condition is satisfied then shade the region towards that point otherwise away from the point.

Let the point be $\left(0,0\right)$.

Now substitute that point in the inequality $y\le -2x+15$.

Therefore, it can be obtained that:

 $\begin{array}{c}y\le -2x+15\\ 0\le -2\left(0\right)+15\\ 0\le 15\end{array}$

As, the condition obtained is true, therefore to draw the inequality $y\le -2x+15$, shade the region towards the point $\left(0,0\right)$.

Therefore, the graph of the inequality $y\le -2x+15$ is:

Step 4 ­- Description of step.

Now draw all the inequalities in one graph.

The graph having all the inequalities is:

Step 5 ­- Description of step.

Now shade the common region.

The graph showing the common region is:

The region shaded in yellow colour is the common region satisfying all the inequalities.

Therefore, the shaded region is the feasible region.

Therefore, the graph of the system of inequalities is:

Step 6 ­- Description of step.

Label the vertices of the feasible region.

The graph showing the labelled vertices of the feasible region is:

Therefore, the coordinates of the vertices of the feasible region are: 

$A\left(1,3\right)$, $B\left(6,3\right)$ and $C\left(4.333,6.333\right)$

Step 7 ­- Description of step.

The given function is: $f\left(x,y\right)=2x+3y$.

Now calculate the values of the given function at the points $A$, $B$ and $C$.

Therefore, it can be obtained that:

At point A, $f\left(1,3\right)=2\left(1\right)+3\left(3\right)=2+9=11$

At point B, $f\left(6,3\right)=2\left(6\right)+3\left(3\right)=12+9=21$

At point C, $f\left(4.333,6.333\right)=2\left(4.333\right)+3\left(6.333\right)=8.666+18.999=27.665$

Step 8 ­- Description of step.

The values of the function at the points A, B and C are 11, 21 and $27.665$ respectively.

Therefore, it can be noticed that the maximum value of the function is $27.665$ at the point $C\left(4.333,6.333\right)$ and the minimum value of the function is 11 at the point $A\left(1,3\right)$.