+0

# linear programming

+3
312
4

Aaron owns a shipping company. He plans to move into his new office which is near to the city Centre. He needs some filing cabinets to organize his files. Cabinet x which costs RM 100 per unit requires 0.6 square meters of the floor space and can hold 0.8 cubic meters of files. Cabinet y which costs RM 200 per unit, requires 0.8 square meters o the floor space and can hold 1.2 cubic meters of files  the ratio of the number of cabin x to the number of cabinet y is not less than 2:3 . Aaron has an allocation of RM1400 for the cabinets and the office has room for no more than 7.2 square meters

1. Write the inequalities which satisfy all the above constraints

2.using two different methods, find the maximum storage volume

Guest Jun 6, 2015

#2
+81090
+13

Let x be the number of  "X" cabinets and y be the number of "Y" cabinets.

And we are told the following.......

x / y  ≥ 2/3 →  3x  ≥ 2y  →  y ≤ ( 3/2 ) x

100x + 200y ≤ 1400  .....  this is the cost constraint

.6x + .8y ≤ 7.2   .......this is the  constraint on the square meters

We also need two more contraints:  x ≥ 0  and y  ≥ 0,   since we can't have a negative number of cabinets!!!

And  we want to  maximize the cubic meters of  file storage .......we can just call this.....   .8x + 1.2y

Have a look at the graph of the inequalities, here.........https://www.desmos.com/calculator/rxzwqo3dnw

The maximum for the objective function occurs at a corner point in the feasible region......the graph shows that there are two "whole number" corner points at  (8, 3)  and (12, 0)....another corner point occurs at (3.5, 5.25)....but.....we can't buy "partial" numbers of cabinets.....!!!

Notice, at (8, 3),   the objective function = .8(8) + 1.2(3)  = 10

At (12, 0), the objective function  = .8(12) + 1.2(0)  = 9.6

It looks like the best option for maximum storage under the given constraints is to purchase 8 of the "X" cabinets and 3 of the "Y" cabinets

Sorry....I don't know a second method.....

CPhill  Jun 7, 2015
Sort:

#1
+91510
0
Melody  Jun 7, 2015
#2
+81090
+13

Let x be the number of  "X" cabinets and y be the number of "Y" cabinets.

And we are told the following.......

x / y  ≥ 2/3 →  3x  ≥ 2y  →  y ≤ ( 3/2 ) x

100x + 200y ≤ 1400  .....  this is the cost constraint

.6x + .8y ≤ 7.2   .......this is the  constraint on the square meters

We also need two more contraints:  x ≥ 0  and y  ≥ 0,   since we can't have a negative number of cabinets!!!

And  we want to  maximize the cubic meters of  file storage .......we can just call this.....   .8x + 1.2y

Have a look at the graph of the inequalities, here.........https://www.desmos.com/calculator/rxzwqo3dnw

The maximum for the objective function occurs at a corner point in the feasible region......the graph shows that there are two "whole number" corner points at  (8, 3)  and (12, 0)....another corner point occurs at (3.5, 5.25)....but.....we can't buy "partial" numbers of cabinets.....!!!

Notice, at (8, 3),   the objective function = .8(8) + 1.2(3)  = 10

At (12, 0), the objective function  = .8(12) + 1.2(0)  = 9.6

It looks like the best option for maximum storage under the given constraints is to purchase 8 of the "X" cabinets and 3 of the "Y" cabinets

Sorry....I don't know a second method.....

CPhill  Jun 7, 2015
#3
+91510
0

Thanks Chris, if I get up to it I would like to take a better look at this question.

It looks like it would take some effort to get ones head around it.

I hope anon says thank you to you :)

Melody  Jun 7, 2015
#4
+3

There is another way-by table or list all possible anwser .

Guest Jun 8, 2015

### 45 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details