A company can use two types of machine A and B, in a munifucturing plant the number of operators required and the running cost per day are gives as cost per day
machine A 6
machine B 3
max available 360
available operators 2 4 280
floor area profit per machine 20 30 write down the inequality constraints and profit function graph the inequalities from the graph determine of machines A and B which should be used maximize profit?
Let x be the number of Machine A and y be the number of Machine B
The total operators needed for Machine A = 2x
The total operators needed for Machine B = 4y
Total available operators = 280
The inequality constraint for operators:
2x + 4y <= 280
The total daily running cost for Machine A = x * 6 = 6x
The total daily running cost for Machine B = y * 3 = 3y
Maximum available running cost = 360
The inequality constraint for running cost:
6x + 3y <= 360
The profit per machine is given as:
Machine A = 20
Machine B = 30
The profit function (which we want to maximize) is:
P = 20x + 30y
Given these inequalities, we can graph them to determine the feasible region where the profit function can be maximized.
We first find the intersection points of the inequalities:
For 2x + 4y = 280, when x = 0, y = 70.
For 2x + 4y = 280, when y = 0, x = 140.
For 6x + 3y = 360, when x = 0, y = 120
For 6x + 3y = 360, when y = 0, x = 60
Now we can plot these points and graph the inequalities on a coordinate plane.
The feasible region is bounded by the following points:
(0,0), (0,70), (60, 0), (40,60)
Now, we will check for the maximum profit at each of these corners in the feasible region:
1. At (0,0): P = 20(0) + 30(0) = 0
2. At (0,70): P = 20(0) + 30(70) = 0 + 2100 = 2100
3. At (60,0): P = 20(60) + 30(0) = 1200 + 0 = 1200
4. At (40,60): P = 20(40) + 30(60) = 800 + 1800 = 2600
From the above results, the maximum profit is achieved at the point (40,60) with a profit of 2600.
Therefore, the company should use 40 machines of type A and 60 machines of type B to maximize the profit.