How Many Buses Maximize Parking Lot Income?

[TonyC]

Killer word problem--spacing

The area of a parking lot is 1590 square meters. A car requires 6 square meters and a bus requires 32 square meters of space. There can be at most 200 vehicles parked at one time. If the cost to park a car is $2.00 and a bus is $6.00, how many buses should be in the lot to maximize income?

PLEASE HELP!

 

[morry]

This sounds like a linear programming q. Have you learned about the simplex algorithm? I would not know how to tackle this without simplex.

 

[TonyC]

No I haven't. What is the algorithm? Maybe I can attempt it using it.

 

[morry]

Google simplex. I would never be able to understand it without going through the lectures.

So a graphical method may be better for you.

What you first must do is write the above stuff as constraints. ie z= 2x +6y (For money)

6x +32y <(or equalto) 1590 and x + y<(or equalto)200. Youll have to graph these two equations (The ones on this line).

So when you have a graph, keep in mind that the optimal solution will lie on one of the vertices. There shouldn't be too many in this case. So plug stuff into the equations and you should get your maximum profit.

Hope that helped.

 

[TonyC]

Can you please break it down further?
I am still perplexed.

How?

 

[morry]

Ok. So youre given all that stuff right? Ill try and explain why I chose those 3 equations.

z= 2x +6y . This is the money equation. The money taken will be $2 for every car and $6 for every bus. Thus money = 2x +6y. Where x and y are the number of cars and buses, respectively.

6x +32y <(or equalto) 1590 This equation is for the space in the car park. Since each car takes up 6 m and a bus 32 and the total space is 1590. So the sum of all the cars and buses must take up less than 1590.

x + y<(or equalto)200 It says that the total number of buses and cars must be les than 200. Hopefully you can start to see what these equations mean.

Then, graph the last two equations as you normally would. Think about the contraints, the solution will be bounded by the lines. The optimal solution will be at the vertex of those two lines.

The graph should look similar to this one: http://www.egwald.com/operationsresearch/images/lpgraph_primal_1.php

Please reply if there's a particular bit you don't get.

 

[TD]

Coïncidentally, the system of the two conditions gives an integer solution:

\left\{ \begin{gathered}<br /> c + b = 200 \hfill \\<br /> 6c + 32b = 1590 \hfill \\ <br /> \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}<br /> c = 185 \hfill \\<br /> b = 15 \hfill \\ <br /> \end{gathered} \right

 

[TonyC]

Now I see...thank you :0

 

[morry]

haha, even easier.

 

[HallsofIvy]

What do you see? Do you know what the answer to this problem is? TD told you that the two equations are both satisfied when the number of cars is 15 and the number of buses is 185. In other words, that is one possible number of cars and buses that will fit on the lot. Do you have any reason to believe that that will give maximum income?


That is, in fact, the correct answer but you be able to prove it. The point of "linear programming" (you don't really need the "symplex method") is that when the set of "feasible" solutions (here the number of cars and buses that you can fit on the lot) is convex polygon and the "object function" (here the income) is linear, then object function takes on a maximum or a minimum only at a vertex of the polygon (imagine moving the straight line graph of the object function parallel to itself {increasing or decreasing its value}. The line will "leave" the polygon at a vertex.).

It's easy to see that the "feasible region" here is a polygon bounded by (taking c for number of cars, b for number of buses) b=0, c= 0; b= 0, c= 200 (where b+ c= 200 crosses the c-axis); c= 0, b= 49.68 (where the 6c+ 32b= 1590 crosses the b-axis);and c= 15, b= 185 (where the two lines cross). Now evaluate the object function 2c+ 6b at each of those to find the maximum value.