Maximize p = x + y
subject to x + 3y <=4
3x + y<= 4
x >=0, y >=0
You would want to plot the 4 lines to make a convex polygon, inside of which is the feasible region, i.e. any point inside the polygon will satisfy the constraints.
Now plot the line Z=x+y, which is a line at 45 degrees sloping downwards to the right. Draw lines parallel to this line until the line touches on the last corner (top right, approximately) of the polygon. The x-y values of this corner is the answer you require.
You can do all this algebraically, but it is a little more abstract. You are better off starting it graphically and then do the algebraic part afterwards to confirm or get better accuracy.
To solve this linear programming problem and maximize the objective function p = x + y, subject to the given constraints, we can follow these steps:
1. Graph the constraints:
- Plot the lines x + 3y = 4 and 3x + y = 4 on a coordinate plane.
- Shade the feasible region that satisfies all the constraints, considering the non-negativity constraints x >= 0 and y >= 0.
2. Identify the corner points of the feasible region:
- The corner points are the intersection points of the lines representing the constraints.
- Calculate the coordinates of each corner point by solving the system of equations formed by the intersecting lines.
3. Evaluate the objective function p = x + y at each corner point:
- Substitute the x and y values of each corner point into the objective function to find the corresponding value of p.
4. Determine the maximum value of p and the corresponding values of x and y:
- Compare the values of p obtained from step 3 and identify the highest value.
- The values of x and y corresponding to this highest value of p will give the solution to the problem.
In this case, the feasible region is a quadrilateral formed by the lines x + 3y = 4, 3x + y = 4, x = 0, and y = 0.
To find the corner points, solve the system of equations:
x + 3y = 4
3x + y = 4
Solving this system, we find two corner points: (0, 4/3) and (4/3, 0).
Evaluate the objective function p = x + y at each corner point:
For (0, 4/3):
p = 0 + 4/3 = 4/3
For (4/3, 0):
p = 4/3 + 0 = 4/3
Hence, the maximum value of p is 4/3, which is achieved at the corner points (0, 4/3) and (4/3, 0).