Linear Programming

4 Applying Linear Programming

Having learned how to use your fx-991ES PLUS

calculator to perform typical linear programming

tasks, it’s time to look at actual applications.




Example 4.1

Planet-15 makes green and
blue widgets. The cost of

production is divided between

machine time (in hours) and
materials and are represented in the following table.


Machine Materials
time
Blue £0.3 £0.25

Green £0.4 £0.5



The total amount of money for machine time is £1.15

a minute. While the money for materials is limited to
£1.25 per item. Normally the profit on the blue

widgets is £0.7 whereas the profit on the green widget
is £1.00. How many widgets of both colours should

Planet-15 make to maximise their profits?

45

Let x be the number of blue widgets and y the number

of green widgets. The constrains are,

For machine time 0.3 + 0.4 ≤ 1.15


For materials 0.25 + 0.5 ≤ 1.25

The objective function is = 7 + 10


From the plots on page 4, this is a 2Type 2 plot. Use the

following keystrokes for your fx-991ES calculator –
first set it up so that MatA is a 2 × 2, matrix, MatB is a

2 × 1 matrix and MatC a 1 × 2 matrix,

(MODE)615Cq(MATRIX)226C

q(MATRIX)23R2C


Scale up the given the constraint inequalities,


30 + 40 ≤ 115
25 + 50 ≤ 125

The keystrokes for MatA,

q(MATRIX)21

30=40=25=50=









46

The keystrokes for MatB,


Cq(MATRIX)22115=125







The keystrokes for MatC for the objective function,

Cq(MATRIX)23 7=10=








Perform the calculation to determine the intersection,
Cq(MATRIX)3uq(MATRIX)4=n







The profit every minute for the objective function
passing through this intersection is,


Cq(MATRIX)5Oq(MATRIX)6=












47

Scaled up, Planet-15 should be making 150 blue

widgets and 175 green widgets every 100 minutes
which will give them a profit of £280. A plot of the

inequalities and the objective function (in green) is

shown below.

































From the graph, objective function for


(0, 2.5) = 7 × 0 + 10 × 2.5 = 25

(3.8, 0) = 7 × 3.8 + 10 × 0 = 26.6





48

Example 4.2

A small company Mantex
manufactures two types of

widget made from a blend of

two different alloys A and B. Below is a table showing
the weight of each element (measured in grams)

needed for each coloured widget.


Alloy A Alloy B
Green 2 3

Yellow 4 2



There is a limited supply each day of the alloys, 360

grams of Alloy A and 270 grams of Alloy B. To fulfil
contracts, Mantex must use 720 grams of alloy each

day. Represent the daily production green widgets as
x and yellow widgets as y. The profit on the green

widget is £2.26 and the profit on each yellow widget

is £3.05. How many of each widget should be made
each day to maximise the profit?





We can infer from the table; looking at column for
Alloy A that,

49

2 + 4 ≤ 360

From column 2


3 + 2 ≤ 270


and the objective function is

= 2.26 + 3.05


Looking at the inequalities, this is a 2Type2 plot. Load
the three matrices in your fx-991ES calculator using

the following keystrokes.

Setting up the matrices: (MODE)6



MatA
15 2=4=3=2C

MatB
q(MATRIX)226

360=270=C

MatC

q(MATRIX)23R2

2.26=3.05=


To find the coordinate of the intersection of the two
inequalities, enter the following keystrokes,




50

Cq(MATRIX)3uq(MATRIX)4=







The objective function passing through this

coordinate is given by,

Cq(MATRIX)5Oq(MATRIX)6=








The profit per day is £307.57 when producing 45

green widgets and 68 yellow widgets.
















(0,90) = 0 × 2.26 + 90 × 3.05 = £274

(90,0) = 90 × 2.26 + 90 × 0 = £203
The optimum solution is the green line shown in the

plot.



51

To gain a better understanding of matrices you are

recommended to read the following.













































Available from,


www.CambridgePaperbacks.com




52

There are other ebooks from Cambridge Paperbacks

which may be of interest to you.


















































53

54

For more interesting maths books, visit





www.CambridgePaperbacks.com






Cambridge 

Paperbacks













55