A school organized a book fair and in this book fair a book seller is selling his books under the following rules:
There are three different packages available.
First package contains 2 Islamic books, 2 Science books and 2 Geography books, second package contains 2 Islamic books, 4 Science books and 1 Geography books and third package contains 3 Islamic books, 4 Science books and 5 Geography books. The book fair has a total of 250 Islamic books, 300 Science books, and 270 Geography books. First package makes a profit of Rs. 120, second package makes Rs.100 and third package makes Rs.270 per pack.
How many packs should be made to maximize book fair profits?
What will the profit be?
1
Expert's answer
2021-12-28T16:25:30-0500
Let x1,x2,x3 denote the number of packages of the first, the second and the third kind, respectively. Then we have the following linear restrictions
The total number of Islamic books must be not greater than 250: 2x1+2x2+3x3≤250
The total number of Science books must be not greater than 300: 2x1+4x2+4x3≤300
The total number of Geography books must be not greater than 270: 2x1+x2+5x3≤270
We want to maximize the book fair profit: 120x1+100x2+270x3→max , given that x1,x2,x3 are non-negative integers.
Let’s derive the standard form of this problem. Denote
y1=250−(2x1+2x2+3x3)
y2=300−(2x1+4x2+4x3)
y3=270−(2x1+x2+5x3)
Z=120x1+100x2+270x3
Then we have the linear programming problem in the standard form:
2x1+2x2+3x3+y1+0y2+0y3=250
2x1+4x2+4x3+0y1+y2+0y3=300
2x1+x2+5x3+0y1+0y2+y3=270
Z=120x1+100x2+270x3→max , given that x1,x2,x3,y1,y2,y3 are non-negative integers.
We will solve this problem by the simplex algorithm. We fill the tableau:
Columns y1,y2,y3 represent the basic variables, the corresponding basic feasible solution is x1=x2=x3=0, y1=250,y2=300,y3=270.
Since the first row contains non-negative entries, the obtained solution is not optimal.
Select x3 as an entering variable. Then min{250/3,300/4,270/5}=270/5=54 and thus, y3 is the corresponding leaving variable. In order to annihilate all entries in column x3, excluding the last one, we should add the 4th row with suitable coefficients to the others:
Since the first row contains non-negative entries, the obtained solution is not optimal.
Select x1 as an entering variable. Then min{88/0.8,84/0.4,54/0.4}=88/0.8=110 and thus, y1 is the corresponding leaving variable. In order to annihilate all entries in column x1, excluding 0.8, we should add the 2nd row with suitable coefficients to the others:
Since the first row contains a non-negative entry, the obtained solution is not optimal.
Select x2 as an entering variable. Then min{110/1.75,40/2.5}=40/2.5=16 and thus, y2 is the corresponding leaving variable. Add the 3rd row with suitable coefficients to the others:
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