Posted by Anonymous on Wednesday, November 26, 2008 at 8:49pm.
P=-30x+25y
Subject to 2x+3y>=30
2x+y<=26
-6x+5y<=50
x,y>=0
I need the minimize and maximaze
Responses
* linear programming problem - Damon, Thursday, November 27, 2008 at 1:35pm
One point at the origin, constrained to first quadrant.
first constraint #1
2x+3y>=30
when x = 0, y = 10
when x = 15, y = 0
area above that line is in our region (are you sure your arrow is correct?)
second constraint #2
2x+y<=26
when x = 0, y = 26
when x = 13, y = 0
area below that line is in
third constraint #3
-6x+5y<=50
when x = 0, y = 10
when x = -50/6 or -8 1/3 , y = 0
area below that line is in
sketch a graph. You see that our region of interest is a triangle from (0,10) to the intersection of #2 and #3 down to the intersection of #1 and #2
first #1 and #2
2x+3y=30
2x+y=26
gives
2 y = 4 or y = 2 then x = 12
so (12,2)
now #2 and #3
2x + y=26 -->6 x + 3y = 78
-6x+5y=50
gives
8y = 128
y = 16 then x = 5
so (5,16)
NOW
calculate P at (5,16) and (12,2) and (0,10) and chose the biggest or smallest whichever you want. If P is "profit", you probably want the biggest.
To solve this linear programming problem, you need to find the minimum and maximum values of the objective function P=-30x+25y, subject to the given constraints.
First, let's graph the constraints on a coordinate plane:
1. Constraint: 2x+3y>=30
- When x=0, y=10
- When x=15, y=0
The area above this line is in our region.
2. Constraint: 2x+y<=26
- When x=0, y=26
- When x=13, y=0
The area below this line is in our region.
3. Constraint: -6x+5y<=50
- When x=0, y=10
- When x=-50/6 or -8 1/3, y=0
The area below this line is in our region.
Based on these constraints, we can see that our region of interest is a triangle from (0,10) to the intersection of constraint #2 and constraint #3 down to the intersection of constraint #1 and constraint #2.
To find the intersection points, solve the equations for the corresponding constraint pairs:
1. Constraint #1 and constraint #2:
2x+3y=30
2x+y=26
Solving this system of equations, we get:
2y=4 or y=2
From the second equation, x=12
Therefore, the intersection point is (12,2).
2. Constraint #2 and constraint #3:
2x+y=26
-6x+5y=50
Multiplying the first equation by 3 to eliminate y, we get:
6x+3y=78
Subtracting this equation from the second equation, we get:
-8y=-128
Solving for y, we get y=16
Substituting y=16 into the first equation, we get x=5
Therefore, the intersection point is (5,16).
Now, calculate the value of P at each of the intersection points: (5,16), (12,2), and (0,10). Plug the x and y values into the objective function P=-30x+25y, and compute the corresponding P values.
Choose the either the minimum or maximum P value, depending on what you are looking for (e.g., profit).
To solve this linear programming problem, you need to find the minimum and maximum values of the objective function P=-30x+25y, subject to the given constraints:
1. 2x+3y>=30
2. 2x+y<=26
3. -6x+5y<=50
4. x,y>=0
To graphically represent the feasible region, plot the lines corresponding to each constraint and shade in the region that satisfies all the constraints:
1. Plot the line 2x+3y=30. When x = 0, y = 10, and when x = 15, y = 0. The area above this line is within the feasible region.
2. Plot the line 2x+y=26. When x = 0, y = 26, and when x = 13, y = 0. The area below this line is within the feasible region.
3. Plot the line -6x+5y=50. When x = 0, y = 10, and when x = -50/6 or -8 1/3 , y = 0. The area below this line is within the feasible region.
The feasible region is the triangular area from (0,10) down to the intersection of the second and third constraint lines, and then across to the intersection of the first and second constraint lines. The points (12,2) and (5,16) are the intersections of the lines, which define the vertices of the feasible region.
To find the minimum and maximum values of P, evaluate P at each vertex: (0,10), (12,2), and (5,16). Choose the largest or smallest value depending on whether you're looking for maximum or minimum profit.