Solution of L.P.P.
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Q1
Marks:1Answer:
(2, 0)
Explanation:
Hence maximum value of 3x + 2y is at (2, 0).

Q2Marks:1
Answer:
24.
Explanation:
Change the inequalities into equations and draw the graph of lines, thus we get
the required feasible region.
It is a bounded region, bounded by the vertices A(1,0), B(8,0) andNow by evaluation of the objective function for the vertices of feasible region it is found to be
maximum at (8,0). Hence the solution is Z = 3x_{1} + 2x_{2} = 3 × 8 + 2 × 0 = 24. 
Q3
A vertex of bounded region of inequalities x + 2y 0 and 2x + y 4, x 0 is
Marks:1Answer:
(0, 0).
Explanation:
From the graph of bounded region we can see that (0, 0) is one of the vertex of feasible region. 
Q4
For the constraints of a L.P. problem given by x_{1} + 2x_{2} 2000, x_{1} + x_{2} 1500, x_{2} 600 and x_{1}, x_{2} 0 the point outside the feasible region is
Marks:1Answer:
(2000, 0).
Explanation:
Clearly point (2000, 0) is outside. 
Q5
A point outside the positive region bounded by the inequalities 2x + 3y 6, 5x + 3y 15 and x, y > 0 is
Marks:1Answer:
(0, 5).
Explanation:
From the graph we can see that (0, 5) lies outside the region.