Question - Optimizing Linear Programming Problem
Solution:
The question in the image shows a linear programming problem where the objective is to maximize the function Z = 10X + 12Y subject to three constraints:1. 2X + 3Y ≤ 15002. 3X + 2Y ≤ 15003. X + Y ≤ 600and the non-negativity constraints:X ≥ 0Y ≥ 0To solve this optimization problem, follow these steps:1. **Graph the Constraints:** Plot the lines represented by each constraint on a graph. The area that satisfies all constraints is known as the feasible region.2. **Find the Intersection Points of the Feasible Region:** Calculate the intersection points of the lines forming the boundaries of the feasible region. You'll need to solve the equations where two lines intersect. This includes the intersections with the X and Y axes.3. **Evaluate the Objective Function at each Intersection Point:** That is, substitute the values of X and Y obtained from the intersection points into the objective function Z = 10X + 12Y to find out which point gives the maximum value of Z.Let's calculate the intersection points:For the X-axis (Y=0), using constraint 3 (X + Y ≤ 600):X = 600, Y = 0For the Y-axis (X=0), using constraint 3 (X + Y ≤ 600):Y = 600, X = 0Now let's find intersection of the lines by pairing the constraints:For the first two constraints:2X + 3Y = 15003X + 2Y = 1500You can solve these equations simultaneously or use methods like substitution or elimination to find the values of X and Y.Let's use the elimination method to solve it straightforwardly:Multiply the first equation by 3 and the second equation by 2:6X + 9Y = 45006X + 4Y = 3000Now, subtract the second equation from the first:6X + 9Y - (6X + 4Y) = 4500 - 30005Y = 1500Y = 300Substitute Y = 300 into the first original constraint:2X + 3(300) ≤ 15002X + 900 ≤ 15002X ≤ 600X = 300So we have another intersection point at (300, 300).Now you will need to find the intersection between the last two constraints 3X + 2Y = 1500 and X + Y = 600 in a similar manner, and then you will need to calculate the values of the objective function Z at each of these intersection points, including the intersections with the axes: (600, 0) and (0, 600).The maximum value of Z will determine the optimal solution for the problem. If that maximum occurs at any of the corners of the feasible region, then that point represents the maximum values of X and Y for the problem.
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