2. s two products: 1 and 2. Each unit of each A manufacturing company produce pr
ID: 3375010 • Letter: 2
Question
2. s two products: 1 and 2. Each unit of each A manufacturing company produce product must be processed on machin and raw material 2. The resources usage, sales price per unit and the demarn each product are giving in the following table e 1 and machine 2 and uses raw material T Demand Sales Price Machine 1 Machine 2 Raw material 1 Raw material 2 Productl 450 35 0.6 0.4 Product2 350 40 0.4 0.3 Thus, producing one unit of producti uses 0.6 unit of machine ltime, 0.4 unit of machine 2 time, 2 units of raw material 1, and 1 unit of raw material 2 It costs $5 to purchase each unit of raw material 1, andS 6 to purchase each unit of raw material 2. Unlimited amount of raw material can be purchased. Two hundred units of machine1 time and 300 units of machine 2 time are available. Determine how the company can maximize its profit.Explanation / Answer
Solution
Let number of units to be produced of Product 1 and Product 2 be x and y respectively.
NOTE: Since question specifically says to maximize profits and only relevant figures given are sales price and raw material cost, we will take
profit = sales price - raw material cost …………………………………………………(1)
Given Product 1 requires 2 units of Raw material 1. 1 unit of Raw material 2 costing $5 and $6 respectively, and sales price is 35, vide (1) above, profit = 19.
Similarly, profit per unit for Product 2 is [40 – {(1 x 5) + (2 x 6)}] = 23. Thus, total profit is:
Z = 19x + 23y ……………………………………………………………………………(2)
Since Products 1 and 2 require 0.6 and 0.4 unit Machine 1 respectively, total requirement of Machine 1 is: 0.6x + 0.4y. Given only 200 units of Material 1 is available, we should have:
0.6x + 0.4y ? 200 …………………………………………………………………………(3)
Similarly, with respect to Machine 2: 0.4x + 0.3y ? 300 ………………………………(4)
With respect to demand, we should have: x ? 450 and y ? 350 ………………………..(5)
Given that unlimited amount of raw material can be purchased, there is no constraint on that account.
Since x and y are production quantities, both need to non-negative. Thus, the problem can be formulated as a LPP (Linear Programming Problem) as follows:
Maximize: Z = 19x + 23y subject to
0.6x + 0.4y ? 200
0.4x + 0.3y ? 300
x ? 450
y ? 350
x, y ? 0.
Solution by Graphical Method
0.6x + 0.4y = 200 is a straight line cutting x-axis at 1000/3 and y-axis at 500
0.4x + 0.3y = 300 is a straight line cutting x-axis at 750 and y-axis at 1000
x = 450 is a straight line parallel to y-axis at a distance of 450 units to the right of y-axis.
y = 350 is a straight line parallel to x-axis at a distance of 350 units above the x-axis.
With these lines and the inequality conditions, the feasible region is the polygon OPQRO, where O = (0, 0), P = (0, 350), Q = (93.3, 350), and R = (1000/3, 0).
Substituting these coordinates in the objective function (2), we have:
ZO = 0, ZP = 8050, ZQ = 9823, ZR = 6333.
Clearly, ZQ = 9823 is the maximum. Hence, the optimum solution is to produce
93 units of Product 1 and 350 units of Product 2. Maximum profit is 9823. ANSWER
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