P=-30x+25y
Subject to 2x+3y>=30
2x+y<=26
-6x+5y<=50
x,y>=0
I need the minimize and maximaze
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 determine the minimum and maximum values of the objective function P = -30x + 25y, subject to the given constraints, we need to solve the system of linear inequalities using the graphical method.
First, we will plot the feasible region represented by the system of inequalities:
1. 2x + 3y ≥ 30
2. 2x + y ≤ 26
3. -6x + 5y ≤ 50
4. x ≥ 0, y ≥ 0
Step 1: Solve each inequality for y in terms of x:
1. y ≥ (30 - 2x) / 3
2. y ≤ 26 - 2x
3. y ≤ (6x + 50) / 5
Step 2: Plot the lines corresponding to the equalities:
1. 2x + 3y = 30: This line has the points (0, 10) and (15, 0).
2. 2x + y = 26: This line has the points (0, 26) and (13, 0).
3. -6x + 5y = 50: This line has the points (-10, 0) and (0, 10).
Step 3: Determine the region of interest by shading the feasible region:
- Shade the area above the line 2x + 3y = 30.
- Shade the area below the line 2x + y = 26.
- Shade the area below the line -6x + 5y = 50.
Step 4: Identify the bounded region that satisfies all constraints. This region is the feasible region.
Step 5: To find the minimum and maximum values of P = -30x + 25y, we evaluate the objective function at the corner points of the feasible region (vertices of the bounded region).
By examining the graph, the vertices of the feasible region are A(0,10), B(6,6), C(9,4), and D(5,5).
Evaluate the objective function at each vertex:
P(A) = -30(0) + 25(10) = 250
P(B) = -30(6) + 25(6) = 0
P(C) = -30(9) + 25(4) = -15
P(D) = -30(5) + 25(5) = -25
The minimum value of P is -30, which occurs at point C(9,4).
The maximum value of P is 250, which occurs at point A(0,10).
To find the minimum and maximum values of the objective function P=-30x+25y, subject to the given constraints, we will use a technique called linear programming.
Step 1: Graph the Constraints
We can start by graphing the constraints on a coordinate plane. This will allow us to visualize the feasible region, which represents the area where all constraints are satisfied.
The constraint 2x+3y>=30 can be graphed as a line with a solid boundary and shading below the line because we have an inequality sign greater than or equal to.
The constraint 2x+y<=26 can be graphed as a line with a solid boundary and shading above the line because we have an inequality sign less than or equal to.
The constraint -6x+5y<=50 can be graphed as a line with a solid boundary and shading below the line because we have an inequality sign less than or equal to.
The constraints x,y>=0 indicate that the region of interest is in the first quadrant of the coordinate plane.
Step 2: Find the Intersection Points
Next, we need to find the intersection points of the lines that represent the constraints. These intersection points are the vertices of the feasible region.
Using algebraic methods, we can find the intersection points by solving the systems of equations formed by pairs of constraints.
For example, we can solve the system of equations formed by the first and second constraints:
2x + 3y = 30
2x + y = 26
Subtracting the second equation from the first equation, we get:
2x + 3y - (2x + y) = 30 - 26
2x + 3y - 2x - y = 4y = 4
Simplifying, we find:
4y = 4
y = 1
Substituting the value of y back into one of the equations, we can solve for x:
2x + y = 26
2x + 1 = 26
2x = 25
x = 12.5
So, the intersection point for the first and second constraints is (12.5, 1).
Similarly, we can find the intersection points for other pairs of constraints:
- First and third constraints: (0, 10)
- Second and third constraints: (8, 10)
Step 3: Evaluate the Objective Function at the Intersection Points
Next, we need to evaluate the objective function P=-30x+25y at each of the intersection points.
For example, at the intersection point (12.5, 1):
P = -30(12.5) + 25(1)
P = -375 + 25
P = -350
Similarly, we can evaluate the objective function at the other intersection points:
- (0, 10): P = -30(0) + 25(10) = 250
- (8, 10): P = -30(8) + 25(10) = 50
Step 4: Determine the Minimum and Maximum Values
Finally, we can determine the minimum and maximum values of the objective function.
In this case, the minimum value is -350, which occurs at the intersection point (12.5, 1). The maximum value is 250, which occurs at the intersection point (0, 10).
So, the minimum value of P is -350, and the maximum value of P is 250.