Question

Maximize z = 0.08x + 0.10y, subject to the constraints
x + y ≤ 15000,
x ≥ 3000,
y ≥ 5000.​

Answer 1

Answer:

x>3000.

I hope it will help you

Answer 2

Answer:

The optimal solution is to produce 7500 units of x and 7500 units of y.

Step-by-step explanation:

To solve this linear programming problem, we can use the graphical method:

First, we plot the feasible region defined by the constraints:

The shaded region represents all the possible combinations of x and y that satisfy the constraints. The line x + y = 15000 represents all the combinations that use up exactly 15000 units of the available resources.

Next, we need to find the corner points of the feasible region, which are the vertices of the polygon. These points are the intersections of the constraint lines:

Corner point A: (3000, 12000)

Corner point B: (7500, 7500)

Corner point C: (15000, 0)

We can now evaluate the objective function at each corner point:

At A: z = 0.08(3000) + 0.10(12000) = 1200

At B: z = 0.08(7500) + 0.10(7500) = 1500

At C: z = 0.08(15000) + 0.10(0) = 1200

The maximum value of z is obtained at corner point B, where z = 1500. Therefore, the optimal solution is to produce 7500 units of x and 7500 units of y.

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