Q29. - 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.

$\begin{array}{c}x\ge 2\\ y\ge 1\\ x-2y\ge -4\\ x+y\le 8\\ 2x-y\le 7\\ f(x,y)=x-4y\\ \end{array}$

Step-by-Step Solution

Answer

Co-ordinates of the vertex of the feasible region $ABCDE$  are$A(2,1),B(4,1),C(5,3),D(\frac{20}{3},\frac{4}{3}),E(2,3)$  respectively.

The maximum and minimum value of$f(x,y)=x-4y$  is  $0and-12$

Step-1 –Concept of solving the linear inequalities.

To solve the inequalities we convert the inequalities into linear equations and find the solution of the equations to obtain the graph.

Step-2 – Concept of shading the region

For shading the region, we choose a point. If the point satisfies the inequalities then the shaded region is towards the point. Otherwise, the shaded region is away from the point.

Step-3 –Solving the Inequalities

Given inequalities are

$x\ge 2;y\ge 1;x-2y\ge -4,x+y\le 8,2x-y\le 7$

Their respective linear equations are$x=2;y=1;x-2y=-4;x+y=8;2x-y=7$

The point which satisfies the equation $x=2$ is $\left(2,0\right)$

The point which satisfies the equation $y=1$ is $\left(0,1\right)$.

The point which satisfies the equation$x-2y=-4$ is $\left(-4,0\right)and\left(0,2\right)$

The point which satisfies the equation$x+y=8$is $\left(0,8\right)and\left(8,0\right)$

The point which satisfies the equation $2x-y=7$ is $\left(\frac{7}{2},0\right)and\left(0,-7\right)$

Step-4- Evaluation of the shaded region

We choose the point$\left(0,0\right)$ to obtain the shaded region.

The point $\left(0,0\right)$ satisfies the inequalities $x\ge 2;y\ge 1;x-2y\ge -4;x+y\le 8;2x-y\le 7$

Step-5- Plotting of graphs

So, the graph of respective inequalities is

The feasible region is$ABCDE$   where co-ordinates of  $A,B,C,D,E$ are  $(2,1),(4,1),(5,3),(4,4),(2,3)$ respectively

Step-6- Determination of maximum and minimum value

$f(x,y)=x-4y$

At point $A(2,1)$

$\begin{array}{c}f(x,y)=2-4\times 1\\ =-2\end{array}$

At point $B(4,1)$

$\begin{array}{c}f(x,y)=4-4\times 1\\ =0\end{array}$

At point $C(5,3)$

$\begin{array}{c}f(x,y)=5-4\times 3\\ =-7\end{array}$

At point $D(4,4)$

$\begin{array}{c}f(x,y)=4-16\\ =-12\end{array}$

At point $E(2,3)$

$\begin{array}{c}f(x,y)=2-12\\ =-10\end{array}$