Task Mathematics Problem Solving 2022

Task 4: Problem Solving
Question 1
(a) The problem obeys proportionality assumption of LP because a change in the cost of
any product (A or B) will have a proportional change in the total cost. The total cost
of production of the beverage is the sum of the cost contributed by product A and B
(additive assumption). The implication is that there is no interaction between the
decision variables. Next, the decision variables (liters of orange, lime and mango) are
continuous they can take both decimal and integers. [1] Cost per unit of product A and
B are known thus satisfying the certainty assumption of LP. Finally, the factory
cannot produce less than zero units of beverages implying that the assumption of
finite choices is satisfied.
(b) The LP is formulated as follows:
Let n = volume in liters of product A and,
m = volume in liters of product B.
The objective function is to minimize the total cost function
f (n, m) = 5n + 7m (1)
The constraints are:
Liters of orange not less than 4.5 liters per 100liters of beverage
6 n+4 m
100 ≥ 4.5 (2)
Liters of mango not less than 5 liters per 100liters of beverage
4 n+6 m
100 ≥5 (3)
Liters of Lime no more than 6 liters per 100liters of beverage
3 n+8 m
100 ≤ 6 (4)

Customer Needs
n+ m≥ 80 (5)
Certainty constraints
n ≥ 0 (6)
m ≥0 (7)
Equation (1) represents the objective function while equation (2) to (7) represent
constraints.
(c) In order to plot the equations, we make m the subject in the constraints as follows:
Equation (2) becomes
m ≥− 3
2 n+ 225
2
Equation (3) becomes
m ≥− 2
3 n+ 250
3
Equation (4) becomes
m ≤− 3
8 n+75
Equation (5) becomes
m ≥−n+80 . Now we can use these equations to plot the graph in desmons.com
calculator [9].

From the plot n = x and y = m. The extreem points within the solution region are
(33.33, 62.5), (35, 60), (125, 0) and (200, 0). We noe substitute these values in
equation (1) to obtain total costs at these points.
For (m = 33.33, n = 62.5) the total cost is TC =5 ( 33.33 ) +7 ( 62.5 )=$ 604.15
For (m = 35, n = 60) the total cost is TC =5 ( 35 ) +7 ( 60 )=$ 595
For (m = 125, n =0) the total cost is TC =5 ( 125 ) +7 ( 0 )=$ 625
For (m = 200, n = 0) the total cost is TC =5 ( 200 ) +7 ( 0 ) =$ 1000
The minimum cost solution is (35, 60) which means that the factory should use 35
liters of product A and 60 liters of product B in making the beaverage.
(d) The excell below shows the range of values of liters of product A that can be used
without the total cost changing from $595 to $625.
The range for the cost ($) of A that can be changed without affecting the optimum

solution obtained in (c) is $ 5 ≤ A ≤ $ 5.85.
Question 2
(a) Let xij ≥ 0 be a decision variable that denotes the number of tons of products j for
j ε { 1=Spring;2=Autumn ; 3=Winter } to be produced from Materials i 2 for
i ε {C=Cotton ,W =Wool , S=Silk } . Also, let π=TR−TC be the profit function to be
maximized. Where TR – total revenue (sales price x demand), and TC – total cost
(production cost (PC) + purchase cost (VC)).
From the information provided:
TR=60 ( xc 1 + xw 1+ xs 1 ) +55 ( xc 2+ xw2 + xs 2 ) +60 ( xc 3+ xw 3 + xs 3 ) which simplifies to:
TR=60 xc 1 +60 xw1 +60 xs 1 +55 xc 2 +55 xw2 +55 xs 2+ 60 xc 3 +60 xw 3+60 xs 3
PC =5 ( xc 1 + xw 1+ xs 1 ) + 4 ( xc 2+ xw 2 + xs 2 ) +5 ( xc3 + xw 3 + xs 3 )
VC=30 xc 1 +45 xw 1+ 50 xs 1 +30 xc2 + 45 xw 2 +50 xs 2+ 30 xc 3 +45 xw 3+ 50 xs 3
TC=35 xc1 +50 xw1 +55 xs 1 +34 xc2 + 49 xw 2 +54 xs 2 +35 xc3 +50 xw 3+55 xs 3
Then the objective function is to maximize
π=25 xc 1+10 xw 1+ 5 x s1 +21 xc2 +6 xw 2 + xs 2+ 25 xc 3 +10 xw 3 +5 xs 3 (1)
Subject to the following constraints:
Demand
xc 1+ xw 1 + xs 1 ≤ 4500 (2)
xc 2+ xw 2 + xs 2 ≤ 3000 (3)
xc 3+ xw 3 + xs 3 ≤3500 (4)
Proportion of Cotton in Spring
xc1
xc1 + xw 1+ xs 1
≥ 0.5 simplifying to xc 1−xw1 −xs 1 ≥ 0 (5)
Proportion of Wool in Spring