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

Please help me solve this!!

Guest Jun 6, 2015

#2**+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

#2**+13 **

Best Answer

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