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Question

Maximize Z = 3x + 2y, subject to constraints are x+2y10, 3x+y15 and x, y 0.


Solution

Our problem is to maximize,

Z = 3x + 2y         .......(i)

Subject to constraints x + 2y 10   ....(ii)

3x + y 15              ....(iii)

x 0, y 0     .....(iv)

Firstly, draw the graph of the line x + 2y = 10

x010y50

Putting (0, 0) in the inequality x + 2y 10, we have

0+2×0  10 0 10           (which is true)

So, the half plane is towards the origin. Since, x, y 0

So, the feasible region lies in the first quadrant.

Secondly, draw the graph of the line 3x+y=15

x05y150

Putting (0, 0) in the inequality 3x + y 15, we have 3×0+015015 (which is true)

So, the half plane is towards the origin. 

On solving equations x+2y=10 and 3x+y=15,

we get x = 4 and y = 3

Intersection point B is (4, 3).

Feasible region is OABCO. The corner points of the feasible region are O(0, 0), A(5, 0), B(4, 3) and C(0, 5). The values of Z at these points are as follows:

Corner pointZ=3x+2yO(0, 0)0A(5, 0)15B(4, 3)18MaximumC(0, 5)10

Therefore, the maximum value of Z is 18 at the point B(4, 3).


Mathematics

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