Use graphical methods to solve tje following linear programming problem.
Minimize: z=3x+2y
subject to: x+y> or = to13
2x+2y> or = to4
X> or = to0, y> or = to0
The minimum value of z is _ , at the corner point _.
I see a problem already, since the two lines are parallel.
They will not intersect to provide a vertex ...
To solve this linear programming problem using graphical methods, we will follow these steps:
Step 1: Graph the feasible region.
Step 2: Identify the corner points of the feasible region.
Step 3: Evaluate the objective function at each corner point.
Step 4: Find the corner point with the minimum objective function value.
Let's start with step 1:
Step 1: Graph the feasible region.
To graph the feasible region, we will plot the lines representing the constraints and shade the region that satisfies all the constraints.
The first constraint is x + y ≥ 13, which is a straight line. To graph this, we can rewrite it as y ≥ -x + 13.
The second constraint is 2x + 2y ≥ 4, which simplifies to x + y ≥ 2.
We can graph these two lines on a coordinate plane:
1) Graph the line y = -x + 13.
- Start by plotting two points: (0, 13) and (13, 0).
- Connect these points with a straight line.
- Shade the region above or on the line. (This represents y ≥ -x + 13)
2) Graph the line y = 2 - x.
- Start by plotting two points: (0, 2) and (2, 0).
- Connect these points with a straight line.
- Shade the region above or on the line. (This represents y ≥ -x + 2)
The intersection of the shaded regions from the two lines represents the feasible region.
Step 2: Identify the corner points of the feasible region.
By examining the graph, we can identify the corner points as the vertices of the shaded region. In this case, the corner points are A(0, 13), B(2, 11), C(4, 9), D(7, 6), E(13, 0).
Step 3: Evaluate the objective function at each corner point.
We need to evaluate the objective function z = 3x + 2y at each corner point.
For point A(0,13), z = 3(0) + 2(13) = 26.
For point B(2,11), z = 3(2) + 2(11) = 28.
For point C(4,9), z = 3(4) + 2(9) = 30.
For point D(7,6), z = 3(7) + 2(6) = 33.
For point E(13,0), z = 3(13) + 2(0) = 39.
Step 4: Find the corner point with the minimum objective function value.
We can see that the minimum value of z is 26, which occurs at the corner point A(0,13).
Therefore, the minimum value of z is 26, at the corner point A(0,13).
To solve the linear programming problem graphically, we need to plot the feasible region and identify the corner points. Then, we can evaluate the objective function at each corner point to find the minimum value of z.
Step 1: Plot the constraints
To plot the constraints, we can start by graphing the boundary lines for each constraint equation and shading the region that satisfies the inequality.
For the first constraint equation, x + y ≥ 13, we draw a straight line with a slope of -1 passing through the point (13, 0) and shade the region above this line (since we want x + y to be greater than or equal to 13).
For the second constraint equation, 2x + 2y ≥ 4, we simplify the equation to x + y ≥ 2 and follow the same process as the first constraint, drawing a line with a slope of -1 passing through the point (2, 0) and shading the region above it.
Finally, to satisfy the non-negativity constraint x ≥ 0 and y ≥ 0, we draw the axes and shade the region in the positive quadrant.
Step 2: Identify the corner points
The corner points are the intersection points of the boundary lines. In this case, we have three corner points: A, B, and C.
Step 3: Evaluate the objective function at each corner point
To find the minimum value of z, we substitute the coordinates of each corner point into the objective function z = 3x + 2y and compute the corresponding values.
Corner point A: (0, 13)
z = 3(0) + 2(13) = 26
Corner point B: (2, 0)
z = 3(2) + 2(0) = 6
Corner point C: (4, 0)
z = 3(4) + 2(0) = 12
Step 4: Find the minimum value of z
Comparing the values obtained at each corner point, we can see that the minimum value of z is 6 at the corner point (2, 0).
Therefore, the minimum value of z is 6, at the corner point (2, 0).