Question 1170479
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If to re-write the problem in a correct way, fixing all the deficiencies 
of the original text, we will get something like THIS:


<pre>
    the officers of a high school senior class are planning to rent buses and vans for a class trip. 
    Each bus can transport 56 students, requires two chaperones, and cost $1200 to rent. 
    Each van can transport eight students, requires one chaperone and cost $120 to rent. 
    Since there are 504 students in the senior class that may be eligible to go on a trip, 
    the officers plan to accommodate at least 504 students. 
    Since only 28 parents having volunteer to serve as chevrons, the officers 
    must plan to use at most 28 chaperones. 
    How many vehicles of each type should the officers order to minimize the transportation cost? 
    What are the minimal transportation cost?
</pre>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>S O L U T I O N</U>


<pre>
It is a standard linear minimization problem. 
Let' solve it using the Linear Programming method. 


Let X be the number of buses and Y be the number of vans,


Then from the problem, we have these constraints

    56*X + 8*Y => 504,     (1)

     2*X + 1*Y <=  28,     (2)

       X >= 0,  Y >= 0.    (3)


The objective function to minimize is

    F(X,Y) = 1200X + 120Y.    (4)



We simplify it to this form:

    restrictions

        7X + Y >= 63    (1')

        2X + Y <= 28    (2')

         X >= 0,  Y >= 0.    (3')


     Objective function to minimize

         F(X,Y) = 1200X + 120Y.    (4')


The feasibility domain is the domain in quadrant QI 
above or on the line 7X + Y = 63 and below or on the line 2X + Y = 28.


Thus the feasibility domain is the triangle with vertices

    P1 = (X,Y) = (9,0)   (X-intercept of the line 7X + Y = 63);

    P2 = (X,Y) = (14,0)  (X-intercept of the line 2X + Y = 28);

    P3 = (X,Y) = (7,14)  (intersection of the lines  7X + Y = 63  and  2X + Y = 28).


According to the LP-method, the solution point is the corner, where the objective function is minimal.


So, we calculate the objective function  F(X,Y) = 1200X + 120Y  in three points P1, P2 and P3

    P1:  F(9,0)  = 1200*9  + 120*0 = 10800;

    P2:  F(14,0) = 1200*14 + 120*0 = 16800;

    P3:  F(7,14) = 1200*7 + 120*14 = 10080.


The minimum value of the objective function is at P3, so the point P3 = (7,14)  provides the optimal solution.


<U>ANSWER</U>.  7 buses and 14 vans provide the optimal solution with the minimum transportation cost of $10,080.
</pre>

Solved.