SOLUTION: A transport company has two types of trucks, Type A and Type B. Type A has a refrigerated capacity of 20m and non refrigerated capacity of 10m while type B has the same overall vol

The better (and the best) solution is (1 type A truck and 266 type B trucks).

It provides 20*1 + 15*266 = 4010 m^3 of the total refrigerated capacity (which is much more than the necessary 3000m^3)

       and  10*1 + 15*266 = 4000 m^3 of the total non-refrigerator capacity (which exactly equals to the necessary 4000 m^3)

       at the price 30*1 + 40*266 = #10670.



While the solution by @amoresroy (0 type A trucks and 277 type B trucks) 

   provides 20*0 + 15*267 = 4010 m^3 of the total refrigerated capacity (which is much more than the necessary 3000m^3)

       and  10*0 + 15*267 = 4005 m^3 of the total non-refrigerator capacity (which is more than necessary 4000 m^3)

       at the GREATER price 30*0 + 40*267 = #10680.

First, clearly, it is a Linear programming problem.

If we are going to solve it as a Linear programming problem, we should provide
the constraints, the objective function and the feasibility domain.


If X and Y be the numbers of type A trucks and type B trucks, then

     the objective function is  P(X,Y) = 30X + 40Y;     (1)

     The constraints are  20X + 15Y >= 3000             (2)
                          10X + 15Y >= 4000             (3)
                          X >= 0, Y >= 0.               (4)

     The feasibility domain is the area in first quadrant above the lines 

         20X + 15Y = 3000,    (5)
         10X + 15Y = 4000.    (6)


If you draw these lines, you will see that that they do not intersect in QI: 
line (6) is everywhere ABOVE line (5) in QI.

Thus the feasibility domain is the part of QI above line (6).

And then it is clear that the objection function should be minimal at either x- or y-intercepts
of line (6). 

After that, an easy check shows that linear function is minimal at y-intercept (X,Y) = (0, 266 2/3).


    +---------------------------------------------------------------------+
    |   B U T (!)   - but we are looking for  integer numbers of trucks   |
    +---------------------------------------------------------------------+
 

It means that the solution X = 0,  Y = 266 2/3  DOES NOT work.


It's all about that the solution must be in integer numbers.


So, our problem is NOT a standard Linear programming.


It belongs to a special sub-type of LP-problems - - - to the type of so called "integer LP-problems".

Their peculiarity is that they require specific methods of analysis.


More concretely and more specific, we should consider the points (X,Y) with integer coordinates X and Y
in the feasibility domain ( i.e. above the line (6) ), that are close to this line.

These points will satisfy all necessary restrictions, and we should check and find a point (or points)
with minimal value of the objective function.


I did this job manually (my MS Excel software helped me). The results are presented in the table below.


 Point   X       Y         non-refrigerated   objective function,
   #    coordinates          volume, m^3        dollars
----------------------------------------------------------
   1	 0	267		4005		10680
   2	 1	266		4000		10670    <<<---=== the optimal solution.
   3	 2	266		4010		10700
   4	 3	265		4005		10690
   5	 4	264		4000		10680
   6	 5	264		4010		10710
   7	 6	263		4005		10700
   8	 7	262		4000		10690
   9	 8	262		4010		10720
  10	 9	261		4005		10710
  11	10	260		4000		10700
  12	11	260		4010		10730

As you see from the table, the optimal solution is 1 truck A and 266 trucks B,
with the total cost of 10670 dollars.


In the table, I analyzed only 10 integer points in the feasibility domain, that are in vicinity of (0, 266 2/3), 
but the tendency is just clear and allows to make a necessary conclusion.