SOLUTION: Consider the following linear program: Max 1A+2B s.t 1A ≤5 1B≤5 2A+2B=12 A,B≥0 a.Show the feasible region b.What are the extreme points of the feasible

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Consider the following linear program: Max 1A+2B s.t 1A ≤5 1B≤5 2A+2B=12 A,B≥0 a.Show the feasible region b.What are the extreme points of the feasible       Log On


   



Question 1065033: Consider the following linear program:
Max 1A+2B
s.t
1A ≤5
1B≤5
2A+2B=12
A,B≥0
a.Show the feasible region
b.What are the extreme points of the feasible region?
c.Find the optimal solution using the graphical procedure.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
to use online graphing software, let x = A and let y = B.
this changes your inequalities and equations as follows:

maximize x + 2y, such that:
x <= 5
y <= 5
2x + 2y = 12
x,y >= 0

you would graph the equations.

those would be:

x = 5
y = 5
2x + 2y = 12
x = 0
y = 0

you would then find the region of feasibility by looking at the areas on the graph that satisfy the inequalities and equations.

those would be:

x >= 5
y >= 5
2x + 2y = 12 (this is an equation and not an inqquality)
x >= 0
y >= 0

your graph will look like this:

$$$

you can see that your feasible region is the area on the graph that is:

on or below the line y = 5
on or to the left of the line x = 5
on or above the line y = 0
on or to the right of the line x = 0
on the line 2x + 2y = 12

what this says is that your solution has to:

be on the line of 2x + 2y = 12
have an x-coordinate that is greater than or equal to 0 and less than or equal to 5.
have a y-coordinate that is greater than or equal to 0 and less than or equal to 5.

your max/min balue should be at the corner points of the feasible region.

those corner points would be at (x,y) = (1,5) and (x,y) = (5,1)

i plotted some other points on the line just to show you that the max/min value is indeed on the corner points of the feasible region.

the values i got for x + 2y at the following coordinates are:

(1,5) = 11
(2,4) = 10
(3,3) = 9
(4,2) = 8
(5.1) = 7

your maximum solution is at the coordinate of (1,5).

all of your constraints have to be satisfied.

at the point (1,5):

x <= 5 is satisfied because x = 1
y <= 5 is satisfied because y = 5
x >= 0 is satisfied because x = 1
y >= 0 is satisfied because y = 5
2x + 2y = 12 is satisfied because 2*1 + 2*5 = 2 + 10 = 12

your solution is that the maximum value is when x = 1 and y = 5.

that is one of the corner points of the feasible region.