what is tye minimum value for P=x-4y over the feasibility region defined by the constraints shown above?

Answer:
The correct option is: -30
Step-by-step explanation:
According to the given graph, the the leftmost vertex of the feasibility region is the intersecting point of lines [tex]y=0[/tex] and [tex]y= 3x+2[/tex]
Now, solving the above two equations, we will get.......
[tex]0=3x+2\\ \\ 3x=-2\\ \\ x=-\frac{2}{3}[/tex]
So, the co ordinate of the leftmost vertex will be: [tex](-\frac{2}{3},0)[/tex]
From the given graph, the other vertices are: [tex](8,0), (8,2)[/tex] and [tex](2,8)[/tex]
Given objective function is: [tex]P= x-4y[/tex]
Now, we need to find the value of that objective function at each vertex. So....
For [tex](-\frac{2}{3},0)[/tex] , [tex]P= -\frac{2}{3}-4(0)=- \frac{2}{3} \approx -0.67[/tex]
For [tex](8,0)[/tex] , [tex]P=8-4(0)=8[/tex]
For [tex](8,2)[/tex] , [tex]P=8-4(2)=8-8=0[/tex]
For [tex](2,8)[/tex] , [tex]P=2-4(8)=2-32=-30[/tex] (Minimum)
Thus, the minimum value of the objective function will be -30.