Q3. - 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

 $\begin{array}{c}3.y\ge 2\\ x\ge 1\\ x+2y\le 9\\ f\left(x,y\right)=2x-3y\end{array}$

Step-by-Step Solution

Answer

The graph of the feasible region is provided below. The vertices of a feasible region are $\left(1,4\right),\left(5,2\right)\text{ and }\left(1,2\right)$.The maximum value of the function is $4\text{ at }\left(5,2\right)$ and minimum value of the function is $-10\text{ at }\left(1,4\right)$.

Step-1 – Apply the concept of graphing the inequality

The steps to graph the inequality are provided below.

1. If the inequality contains greater than or less than sign then the boundary of the line will be dashed. If the inequality contains signs of greater than or equal to or less than or equal to then the boundary of the line will be solid. 

2. Select a point (known as test point) from the plane that does not lie on the boundary on the line and substitute it in the inequality. 

3. If the inequality is true then shade the region that contains the test point otherwise shade the other region when inequality is false.

Step-2 – Apply the concept of linear programming

Linear programming is a technique to find the maximum and the minimum value of a given function over a given system of some inequalities, with each inequality representing a constraint. Graph the inequalities and obtain the vertices of the feasible region (solution set). Substitute the coordinates of the feasible region in the function and determine the maximum and the minimum value.

Step-3 – Graph the inequalities

Graph the inequalities $y\ge 2,x\ge 1\text{ and }x+2y\le 9$ on same plane and shade the region.

The corresponding equation is $y=2$.

Take a test point that does not lie on the boundary of the line, say $\left(1,0\right)$. 

Substitute the point $\left(1,0\right)$ in the inequality and check whether it’s true or not.

$0\ge 2$

This is false.

Therefore, shade the region not containing the point $\left(1,0\right)$.

The corresponding equation is $x=1$.

Take a test point that does not lie on the boundary of the line, say $\left(0,0\right)$. 

Substitute the point $\left(0,0\right)$ in the inequality and check whether it’s true or not.

$0\ge 1$

This is false.

Therefore, shade the region not containing the point $\left(1,0\right)$.

The corresponding equation is $x+2y=9$.

Take a test point that does not lie on the boundary of the line, say $\left(1,0\right)$. 

Substitute the point $\left(1,0\right)$ in the inequality and check whether it’s true or not.

 $\begin{array}{c}1+2\left(0\right)\le 9\\ 1\le 9\end{array}$

This is true.

Therefore, shade the region containing the point $\left(1,0\right)$.

Now, shade the regions.

In the above graph, red region represents the inequality $y\ge 2$, blue region represents the inequality $x\ge 1$ and green region represents $x+2y\le 9$.

Step-4 – Find the feasible region

The feasible region is the region common to all the inequalities which is represented below.

The coordinates of the vertices of the feasible region are $\left(1,4\right),\left(5,2\right)\text{ and }\left(1,2\right)$.

Step-5 – Find maximum and minimum values

Now, to find the maximum and minimum value of the function, substitute the coordinates of the feasible region in the function and determine the maximum and the minimum value.

So, substitute $\left(1,4\right),\left(5,2\right)\text{ and }\left(1,2\right)$ in the function $f\left(x,y\right)=2x-3y$ and evaluate the maximum and minimum value as,

 $\begin{array}{ccc}\left(x,y\right)& 2x-3y& f\left(x,y\right)\\ \left(1,4\right)& 2\left(1\right)-3\left(4\right)& -10\\ \left(5,2\right)& 2\left(5\right)-3\left(2\right)& 4\\ \left(1,2\right)& 2\left(1\right)-3\left(2\right)& -4\end{array}$

From the above table, it is observed that the maximum value of the function is $4\text{ at }\left(5,2\right)$ and minimum value of the function is $-10\text{ at }\left(1,4\right)$.