Name the vertices and then find the values of x and y that minimize the objective function.
x >= 2
y >= 0
3x + 2y >= 12
Minimum for C = x + 5y
Help with system of constraints please! I'm terrible at this...
take a look at
http://www.zweigmedia.com/RealWorld/LPGrapher/lpg.html
Enter your problem. It will find a solution and even provide a graph. Be sure to set the x- and y-ranges appropriately. You can get an example to show the syntax. For this problem, just copy/paste
minimize c=x+5y subject to
x >= 2
y >= 0
Thanks! You really helped, I was stuck on this forever.
To find the vertices and solve this linear programming problem, we need to graph the system of inequalities and then find the feasible region. The vertices of the feasible region are the points where the boundary lines intersect.
First, let's graph each constraint individually and shade the region that satisfies the inequality:
1) x >= 2:
Draw a vertical line passing through x = 2, and shade to the right of it since x is greater than or equal to 2.
2) y >= 0:
Draw a horizontal line passing through y = 0, and shade above it since y is greater than or equal to 0.
3) 3x + 2y >= 12:
Rearrange the equation to y >= (-3/2)x + 6. Start by plotting the y-intercept (0, 6), then use the slope (-3/2) to draw the line. Shade the region above this line.
Now, the feasible region is the overlapping shaded region that satisfies all the inequalities. The vertices of this region are the points where the boundary lines intersect.
To find the vertices, you can pick any two border lines and solve their intersection points:
- Intersection of x = 2 and y = 0: (2, 0)
- Intersection of x = 2 and y = (-3/2)x + 6: (2, 3)
Now, let's evaluate the objective function C = x + 5y at each vertex to find the minimum value:
For (2, 0):
C = 2 + 5 * 0 = 2
For (2, 3):
C = 2 + 5 * 3 = 17
Therefore, the minimum value for the objective function occurs at (2, 0) with a value of C = 2.
Note: If the feasible region is unbounded, you may not have vertices, and there may not be a minimum/maximum value for the objective function.