Maximize Z = 7x + 10y

Subject to

x+y≤30000

y≤12000

x≥6000

x≥y

x, y≥0

Asked by Sakshi | 1 year ago |  186

##### Solution :-

We have to maximize Z = 7x + 10y
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 30000,y = 12000, x = 6000, x = y, x = 0 and y = 0.

Region represented by x + y ≤ 30000:
The line x + y = 30000 meets the coordinate axes at A(30000, 0)A30000, 0 and B(0, 30000)B0, 30000 respectively. By joining these points we obtain the line x + y = 30000.
Clearly (0,0) satisfies the inequation x + y ≤ 30000. So,the region containing the origin represents the solution set of the inequation x + y ≤ 30000.

The line y = 12000 is the line that passes through C(0,12000) and parallel to x axis.

The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.

Region represented by x ≥ y
The line x = y is the line that passes through origin.The points to the right of the line x = y satisfy the inequation x ≥ y.
Like by taking the point (−12000, 6000).Here, 6000 > −12000 which implies y > x. Hence, the points to the left of the line x = y will not satisfy the given inequation x ≥ y.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

The feasible region determined by the system of constraints, x + y ≤ 30000, y ≤ 12000, x ≥ 6000, x ≥  y , x ≥ 0 and y ≥ 0 are as follows:

The corner points of the feasible region are D(6000, 0), A(3000, 0)A3000, 0, F(18000, 12000)F18000, 12000 and E(12000, 12000)E12000, 12000.

We see that the maximum value of the objective function Z is 246000 which is at F(18000, 12000)F18000, 12000 that means at x = 18000 and y = 12000.
Thus, the optimal value of Z is 246000.

Answered by Aaryan | 1 year ago

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