Question

Identify the vertices of the feasible region and use them to find the maximum and/or minimum value for the given linear programming constraints

Answer 1

To solve this question, we are given an equation:

`z=x+3y`

With the constraints:

`\begin{gathered} y\ge-(2)/(3)x+2 \\ . \\ y\leq(3)/(2)x+2 \\ . \\ y\leq-5x+15 \end{gathered}`

We need to find the vertices of the region to find the values of x.

The next step is to equate the equations by pairs:

`\begin{cases}y={-(2)/(3)}x+2 \\ y={(3)/(2)x+2}\end{cases}`
`-(2)/(3)x+2=(3)/(2)x+2`

And solve:

`\begin{gathered} 2-2=x((3)/(2)+(2)/(3)) \\ . \\ 0=x((3)/(2)+(2)/(3)) \\ . \\ x=0 \end{gathered}`

Now, we can find y:

`y=-(2)/(3)\cdot0+2=2`

This vertex is: (0, 2)

Now, the next pair:

`\begin{gathered} \begin{cases}y={(3)/(2)x+2} \\ y={-5x+15}\end{cases} \\ \end{gathered}`
`\begin{gathered} (3)/(2)x+2=-5x+15 \\ . \\ (3)/(2)x+5x=15-2 \\ . \\ x((3)/(2)+(10)/(2))=13 \\ . \\ (13)/(2)x=13 \\ . \\ x=13\cdot(2)/(13) \\ . \\ x=2 \end{gathered}`

And find the value of y:

`y=-5\cdot2+15=-10+15=5`

The coordinates of this vertex are: (2, 5)

Finally, the last pair:

`\begin{cases}y={-(2)/(3)}x+2 \\ y={-5x+15}\end{cases}`
`\begin{gathered} -(2)/(3)x+2=-5x+15 \\ . \\ -(2)/(3)x+5x=15-2 \\ . \\ x((15)/(3)-(2)/(3))=13 \\ . \\ (13)/(3)x=13 \\ . \\ x=13\cdot(3)/(13)=3 \end{gathered}`

Now find y:

`y=-5\cdot3+15=-15+15=0`

The coordinates of this vertex are: (3, 0)

Now, we need to evaluate z on those vertices to find the maximum and minimum values of z.

First vertex (0, 2):

`z=0+3\cdot2=6`

Second vertex (2, 5):

`z=2+3\cdot5=2+15=17`

Third vertex (3, 0)

`z=3+3\cdot0=3`

The answer is:

Maximum value of z: 17

Minimum value of z: 3