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Published by Cambridge Paperbacks, 2018-12-12 18:49:41

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.













































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