P=20x+30y
subject to:
2x+3y>30
2x+y<26
-6x+5y<50
x,y >0
Two ways:
Graphical...
plot the contraint areas 2x+3y=30
2x+y=26
-6x+5y=50
x,y >0
Now we have a nice theorem that states the max, and min, will occur on the boundry where constraint lines meet. So test the profit function 20x+30y at each corner, and you will find the max.
Second method: Analytical
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=1e692c6f72587b2cbd3e7be018fd8960&title=Linear%20Programming%20Calculator&theme=blue
looks like linear programming.
let's look at each region.
2x + 3y > 30
the boundary is 2x + 3y = 30
when x = 0, y = 10
when y = 0 , x = 15 , so we have (0,10) and (15,0)
plot these points on the axes and draw a dotted line,
shade in the region above this line
2x+y < 26
repeat my method, shade in the region below the line 2x+y = 26
careful with the last one, I will do that one as well.
-6x+y < 50 -----> y < 6x+50
draw the boundary line y = 6x+50
when x=0, y = 50 , when y = 0 x = -25/3
so your two points are (0,50) , (-25/3,0)
shade in the region below that line
Your boundary lines should look like this
http://www.wolframalpha.com/input/?i=plot+2x+%2B+3y+%3D+30,+2x%2By+%3D+26,+y+%3D+6x%2B50
and here the actual intersection of the shaded regions.
http://www.wolframalpha.com/input/?i=plot+2x+%2B+3y%3E+30,+2x%2By+%3C+26,+y+%3C+6x%2B50
notice that P = 20x + 30y has the same slope as 2x + 3y = 30
so the farthest point to the right of your region will be the solution.
Solve the corresponding equations to find that point, plug into
P = 20x + 30y
To solve the given linear programming problem step-by-step, you can follow these steps:
Step 1: Graph the feasible region:
First, we need to graph the inequalities to determine the feasible region. Let's start with each inequality one by one:
2x + 3y > 30:
To graph this inequality, we can consider it as an equation and plot the corresponding line using two points. Let's choose two arbitrary points and find their coordinates when 2x + 3y = 30:
When x = 0, 2(0) + 3y = 30 => 3y = 30 => y = 10 (0, 10)
When y = 0, 2x + 3(0) = 30 => 2x = 30 => x = 15 (15, 0)
Plotting these two points and drawing a line passing through them, we will have a solid line since the inequality is greater than.
2x + y < 26:
Similarly, we can find two arbitrary points on this line:
When x = 0, 2(0) + y = 26 => y = 26 (0, 26)
When y = 0, 2x + 0 = 26 => 2x = 26 => x = 13 (13, 0)
Plotting these points and drawing a line passing through them, we obtain a dashed line since the inequality is less than.
-6x + 5y < 50:
When x = 0, -6(0) + 5y = 50 => 5y = 50 => y = 10 (0, 10)
When y = 0, -6x + 5(0) = 50 => -6x = 50 => x = -8.3 (approximately) (-8.3, 0)
Plotting these points and drawing a line passing through them, we have another dashed line.
Now, we have the three lines representing the given inequalities.
Step 2: Determine the feasible region:
The feasible region is the area that satisfies all the given inequalities. Since we have three inequalities, the feasible region is the intersection of all the shaded areas formed by the three lines.
Step 3: Identify the corner points of the feasible region:
The corner points of the feasible region are the vertices of the polygon formed by the intersection of the shaded areas. We need to find these points by solving the equations formed by the lines intersecting at the vertices.
Step 4: Evaluate the objective function at each corner point:
Plug the x and y-coordinates of each corner point into the objective function P = 20x + 30y to determine the corresponding value of P.
Step 5: Select the optimal solution:
Compare the values of P obtained at each corner point and determine the solution that maximizes or minimizes the objective function, depending on the given problem.
Note: Since we haven't determined the feasible region or corner points yet, I cannot provide specific values for P and the optimal solution. Please follow the steps outlined above to find the answer.
To solve this linear programming problem, we need to find the values of x and y that satisfy the given constraints and optimize the objective function P=20x+30y.
Step 1: Graph the Constraints
To begin, let's graph the given constraints on a coordinate plane.
The first constraint is 2x+3y>30, which can be rewritten as y>(30-2x)/3. To graph this, plot the line y=(30-2x)/3 as a dashed line.
The second constraint is 2x+y<26, which can be rewritten as y<26-2x. To graph this, plot the line y=26-2x as a dashed line.
The third constraint is -6x+5y<50, which can be rewritten as y<(50+6x)/5. To graph this, plot the line y=(50+6x)/5 as a dashed line.
However, we also have the additional constraint that x and y should be greater than zero, which means they are in the first quadrant of the coordinate plane. Shade the region that satisfies this constraint.
Step 2: Determine the Feasible Region
Now, analyze the shaded region in the graph. The feasible region is the area where all the constraints are satisfied simultaneously. Therefore, the feasible region is the intersection of the shaded region and the areas below the dashed lines.
Step 3: Find the Vertices of the Feasible Region
The vertices of the feasible region are the points where two or more of the lines intersect. Identify these points on the graph.
Step 4: Evaluate the Objective Function at the Vertices
Substitute the x and y values of each vertex into the objective function P=20x+30y. Calculate the corresponding P values for each vertex.
Step 5: Determine the Optimal Solution
Compare the P values calculated at each vertex. The optimal solution is the vertex that maximizes or minimizes the objective function, depending on whether we are maximizing or minimizing.
In this case, since P=20x+30y, we are trying to maximize P. So, identify the vertex that yields the highest P value. That vertex represents the optimal solution to the linear programming problem.