  Question

Maximize Z = 3x + 2y, subject to constraints are x+2y≤10, 3x+y≤15 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+0≤15⇒0≤15 (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)18→MaximumC(0, 5)10 Therefore, the maximum value of Z is 18 at the point B(4, 3). Mathematics

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 