Referring to exercise 15, we get the following data:
Let the company manufactures x number of type A sweaters and y number of type B. We make the following table from the given data:
Thus according to the table, the profit becomes, Z=200x+120y
Now, we have to maximize the profit, i.e., maximize Z=200x+120y
The constraints so obtained, i.e., subject to the constraints,
The company spends at most Rs 72000 a day.
∴ 360x + 120y ≤ 72000
Divide throughout by 120, we get
=> 3x+y≤ 600 …(i)
Also, company can make at most 300 sweaters.
∴ x+y≤ 300 …(ii)
Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100
i.e., y-x≤ 100
y≤ 100+x………. (iii)
And x≥0, y≥0 [non-negative constraint]
So, to maximize profit we have to maximize, Z=200x+120y, subject to
Now let us convert the given inequalities into equation.
We obtain the following equation
x ≥ 0
y ≥ 0
The region represented by 3x+y≤ 600:
The line 3x+y=600 meets the coordinate axes (200,0) and (0,600) respectively. We will join these points to obtain the line 3x+y=600. It is clear that (0,0) satisfies the inequation 3x+y≤ 600. So the region that contain the origin represents the solution set of the inequation 3x+y≤ 600
The region represented by x+y≤ 300:
The line x+y=300 meets the coordinate axes (300,0) and (0,300) respectively. We will join these points to obtain the line x+y=300. It is clear that (0,0) satisfies the inequation x+y≤300. So the region that contain the origin represents the solution set of the inequation x+y=300
The region represented by y≤ 100+x:
The line y= 100+x meets the coordinate axes (-100,0) and (0,100) respectively. We will join these points to obtain the line y= 100+x. It is clear that (0,0) satisfies the inequation y≤ 100+x. So the region that contain the origin represents the solution set of the inequation y≤ 100+x
Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations
The graph of these equations is given.
The shaded region OBCDE is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region.
Corner Points are O(0, 0), B(0, 100), C(100, 200), D(150, 150) and E(200,0)
Now we will substitute these values in Z at each of these corner points, we get
So from the above table the maximum value of Z is at point (150,150).
Hence, the maximum profit to the manufacturer is Rs. 48,000, for making 150 sweaters each of type A and type B.
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