Solve the LP using graphical method

Maximize p = 30x + 40y subject to
x + y <= 90
4x + 2y <= 150
50x + 100y <= 4500
x>=0, y>=0
Enter the answers(numbers only) in the order p, x, y

We can't do much graphing here. I am wondering exactly what you want to know? A tutorial?

Secondly, I am wondering why on three simplex posts you put different names on each post? What do you think instructors think when the see multiple questions from the same person, but with different names on them, with NO work or amplification?

hallo

i actually don't understand ua complain
so do you mean to say that i have to first draw the graph to be able to answer the question?
i dnt know the rest

http://personal.jccmi.edu/BaarsonMonaG/Math%20145%20Notes%20for%20Class/11-%20%20Solving%20linear%20programming%20problems%20graphically.pdf

To solve the given linear programming problem using the graphical method, follow these steps:

Step 1: Convert the inequalities into equations:
Convert all the inequalities to equations by replacing the inequality signs with equal signs. Your equations will be:
Equation 1: x + y = 90
Equation 2: 4x + 2y = 150
Equation 3: 50x + 100y = 4500

Step 2: Plot the feasible region:
Graph each equation on a coordinate plane separately. Since we have two variables (x and y), we will use a two-dimensional graph. Plot the equations by finding their respective intercepts.
For Equation 1: x + y = 90
- Plot the points (90, 0) and (0, 90).
- Draw a line passing through these two points.

For Equation 2: 4x + 2y = 150
- Plot the points (37.5, 0) and (0, 75).
- Draw a line passing through these two points.

For Equation 3: 50x + 100y = 4500
- Plot the points (90, 0) and (0, 45).
- Draw a line passing through these two points.

The feasible region of the LP is the intersection of the shaded areas formed by the lines.
Mark the area where all the lines intersect as the feasible region.

Step 3: Identify the corners of the feasible region:
The corners of the feasible region are the points where the lines intersect. Find these points by solving the simultaneous equations of the lines.
In this case, solving the equations will give you the following points:
Corner 1: (0, 45)
Corner 2: (0, 75)
Corner 3: (18, 72)
Corner 4: (30, 60)
Corner 5: (37.5, 52.5)
Corner 6: (45, 45)

Step 4: Calculate the objective function at each corner point:
Evaluate the objective function p = 30x + 40y at each corner point calculated in Step 3. The corner point that gives the maximum value of p is the optimal solution to the LP.

Calculating p at each corner point:
Corner 1: p = 30(0) + 40(45) = 1800
Corner 2: p = 30(0) + 40(75) = 3000
Corner 3: p = 30(18) + 40(72) = 3120
Corner 4: p = 30(30) + 40(60) = 3600
Corner 5: p = 30(37.5) + 40(52.5) = 3900
Corner 6: p = 30(45) + 40(45) = 4050

Step 5: Determine the optimal solution:
The corner point that gives the maximum value of p is the optimal solution to the LP. From Step 4, we can see that the maximum value of p is 4050, which occurs at Corner 6: (45, 45).

Therefore, the answers to the LP are:
p = 4050
x = 45
y = 45