Linear Equations II

Hong Kong

Stage 4 - Stage 5

Lesson

Certain real-world problems arise where, subject to constraints on time and materials, the cost of production (or alternatively the profit produced from production) of manufactured goods and/or services needs to be optimised.

Such problems are referred to as linear programming problems when the constraints can be conceived as linearly varying quantities.

*1. A typical problem*

The best way to examine the linear programming technique is to consider a hypothetical example.

Imagine a recycling company that needs to buy two types of waste bins. The 'Bin-it' bin costs $$100$100 each but the 'Fill-it-up' bin costs twice as much, and the amount available for purchases has to be no more than $$1400$1400. The Bin-it bins take up $6$6 square metres of factory yard space and can hold $8$8 cubic metres of recyclable waste. The more expensive bins take up $8$8 square metres of space but can hold $12$12 cubic metres of waste. The yard is fairly small, confined to $72$72 square metres of useable area. How many of each bin type should you buy to maximise the total recycling storage?

*2. Defining the variables*

The first thing you might notice is that the information in the question is dense, and can be off-putting on a first read. So its a good first strategy to simply re-read the question.

The last sentence provides the question - how many of each type to buy. So we can define two variables, say $x$`x` and $y$`y` as follows:

Let $x$`x` be the number of Bin-it bins

Let $y$`y` be the number of Fill it up bins

*3. Identifying the constraints*

If we had lots of money and lots of space we could buy lots of bins of each type. From what we are told, however, we are *constrained* by both money and space. The next task then is to develop some mathematical expressions that summarise these constraints.

Note first that if we were to buy $x$`x` Bin-it bins, the cost would be $$100x$100`x`, and if we were to buy $y$`y` Fill-it-up bins the cost of those would be $$200y$200`y`. This means that the total cost of $x$`x` Bin-it bins and $y$`y` Fill-it-up bins would be the sum $100x+200y$100`x`+200`y`, and this must not exceed our limit of $$1400$1400. Our first constraint then is given as $100x+200y\le1400$100`x`+200`y`≤1400.

Secondly, our yard space is limited. From what we are told, buying $x$`x` Bin-it bins will take up $6x$6`x` metres of yard space and buying $y$`y` Fill-it-up bins will take up $8y$8`y` metres of yard space. Since there is a maximum space of $72$72 square metres, we can write $6x+8y\le72$6`x`+8`y`≤72.

Finally, we note the obvious that the quantities $x$`x` and $y$`y` cannot be negative numbers. We can write these constraints as $x\ge0$`x`≥0 and $y\ge0$`y`≥0.

So in summary the constraints are listed in the table:

$100x+200y\le1400$100x+200y≤1400 |
(1) |
---|---|

$6x+8y\le72$6x+8y≤72 |
(2) |

$x\ge0$x≥0 |
(3) |

$y\ge0$y≥0 |
(4) |

*4. Identifying the objective function*

We now need to identify what is the objective function. We need to maximise the storage of recycling material. If we call the total storage $S$`S`, then since $x$`x` Bin-it bins store $8x$8`x` cubic metres, and $y$`y` Fill-it-up bins store $12y$12`y` cubic metres, then we can write:

$S=8x+12y$`S`=8`x`+12`y`

Does the test point $\left(0,0\right)$(0,0) satisfy the inequality $-2x-3y\ge6$−2`x`−3`y`≥6?

Yes

ANo

B

What are the coordinates of the point of intersection of the boundary lines in the following system?

$x\ge2$`x`≥2

$y\le-3$`y`≤−3

Give your answer in the form $\left(a,b\right)$(

`a`,`b`).

The graph shows the region of feasible solutions. The objective function is $z=4x+3y$`z`=4`x`+3`y`.

Loading Graph...

What is the value of the function for the point $\left(2,2\right)$(2,2)?

What is the value of the function for the point $\left(3,9\right)$(3,9)?

What is the value of the function for the point $\left(6,10\right)$(6,10)?

What is the value of the function for the point $\left(8,4\right)$(8,4)?

The maximum value of the objective function is $\editable{}$ at the point $($($\editable{}$, $\editable{}$$)$).

The minimum value of the objective function is $\editable{}$ at the point $($($\editable{}$, $\editable{}$$)$).