linear programming

Question

Solve the following linear programming problem, and state what the maximum value is:
 
 Max 11 x+ 5
 

Subject to

12x + 4y <= 48

x + y >= 6

 

x >= 0

y >= 0

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Answer 1

 

First, we need to graph the regions.

 

Treating the inequalities as lines temporarily for the sake of graphing:

 

The first line is:

4y = 48 - 12x 
y = 12 - 3x

The second line is: y = 6 - x

Their intersection is found by setting them equal:
12 - 3x = 6-x

0 = 6 - x - 12 + 3x

0 - 2x - 6

x=3

intersect at (3,3)

Now substitute the origin (0,0) into the inequalities.

 

The first linear equality produces a true statement: 0 <= 12 - 3(0) or 0 <= 12

 

The second linear equality produces a false statement: 0 >= 6 - 0  or 0 >= 6

 

The the shaded region belongs below the first line but above the second.

 

There are three vertices in this graph of the region:

 

(0,4)
(0,12)
(3,3)

 

Plugging these into your max objective function of 11x + 5

 

shows that (3,3) produces the largest value of 38 as opposed to 5 for the other two