Question 1118832
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<pre>
Let X and Y be the numbers of trucks A and B, respectively.


Then the constrains are


X + Y >= 15
2X + 3Y >= 36

X >= 0,  Y >= 0

The cost function is C(X,Y) = 15000*X + 24000*Y to minimize.


The feasibility area is shown in the Figure below



{{{graph( 330, 330, -5.5, 25.5, -5.5, 20.5,
          15-x, (36-2x)/3
)}}}


Plots  X + Y = 15  (red)  and  2X + 3Y = 36 (green).



The feasibility domain is the unbounded area in QI over the green and red lines.


The points to check the cost function are

    P1 = (0,15)    (Y-intercept to red line)

    P2 = (9,6)     (intersection of the red and green lines)

    P3 = (18,0)    (X-intecept to green line).


The rest is just arithmetic.

The values of the cost function are


    at P1:  C(0,15) = 0*15000 + 15*24000 = 360000;

    at P2:  C(9,6)  = 9*15000 + 6*24000  = 279000;

    at P3:  C(18,0) = 18*15000 + 0*24000 = 270000.


You are looking for the minimum - hence, your solution is at P3.


<U>Answer</U>.  18 trucks of the type A and 0 trucks of the type B.
</pre>


Your solution is correct.


Nice job ! !


My congratulations ! ! !


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To see other similar problems solved by the same method, look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Solving-minimax-problems-by--the-Linear-Programming-method.lesson>Solving minimax problems by the Linear Programming method</A> 

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