SOLUTION: Can some one please check my "work" a moving company wants to purchase a min of 15 trucks with a load capacity of at least 36 tons. Model a holds 2 tons and costs 15,000 Model b

Question 1118832: Can some one please check my "work"
a moving company wants to purchase a min of 15 trucks with a load capacity of at least 36 tons. Model a holds 2 tons and costs 15,000 Model b holds 3 toms and costs 24,000 how many of each to minimize costs and what is the cost
Here is what I have so far
M= 15A+24B
A+B=15
B= -A+15
2A+3B _> 36
B= -2/3A + 12
Vertexs
0,15
9,6
18,0
to get lowest cost and min amount of capacity they need to buy 18 "truck A" and 0 "Truck B"
is this correct or am I missing a step?

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.

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 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. Answer. 18 trucks of the type A and 0 trucks of the type B.

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
    - Solving minimax problems by the Linear Programming method
in this site.

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!

If you have several problems like this to solve, let me show you something that might save you some time and effort.

The boundary lines of the constraint functions in the problem are

(1) x+y=15 --> y = -x+15 --> slope = -1
(2) 2x+3y=36 --> y = (-2/3)x+12 --> slope = -2/3

The objective function boundary line is

(3) 15x+24y=a (a is a constant we don't care about) --> y = (-5/8)x+b (another constant we don't care about) --> slope = -3/5

The slope of the objective function is larger (less negative) than the slope of either objective function.

The corner of the feasibility region where the objective function achieves it minimum value is the corner where a line with slope -3/5 just touches the feasibility region without passing through it.

Referring to the diagram in the response provided by tutor @ikleyn, it is easy to see that the corner where that happens is (18,0).

In this problem, with only two constraints, this method only saves you a small amount of work. But in a problem with more constraints and therefore many more corners of the feasibility region, it can save you a lot of time, because you know which corner is going to give you the answer. To get the answer, you will only have to find one of the intersection points of the constraint lines and evaluate the objective function at that one point.
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