Solve the following linear programming problem. Restrict x ≥ 0 and y ≥ 0. Minimize g = 38x + 23y subject to the following.
x + y ≥ 100
−x + y ≤ 20
−2x + 3y ≥ 30
(x, y) =
g =
sketch it
lines a , b and c
above line a
below line b
above line c
looking at it, must be to the right of line a, between the intersections of a,b and a,c
it will be one of those intersections because we are looking for the LEAST value of 38x + 23 y
now find those two intersections
a and b
x + y = 100
-x + y = 20
-----------------------
2 y = 120
y = 60
x = 40 so at (40,60 )
and g would be 38 *40 + 23 * 60
now the other point
-2x + 3 y = 30
2x + 2y = 200
==============
5y = 230
y = 46
x = 54 so (54,46)
and g would be 38*54 + 23 * 46
pick the smaller of the g values at those two points
To solve the linear programming problem, we need to find the values of x and y that minimize the objective function g = 38x + 23y, subject to the given constraints.
Let's start by graphing the constraints:
1. x + y ≥ 100: This is a line with a slope of -1 passing through the point (0, 100) and (100, 0). To graph it, draw a dashed line and shade the region above the line.
2. -x + y ≤ 20: This is a line with a slope of 1 passing through the point (0, 20) and (20, 0). To graph it, draw a dashed line and shade the region below the line.
3. -2x + 3y ≥ 30: This is a line with a slope of 2/3 passing through the point (0, 10) and (15, 0). To graph it, draw a solid line and shade the region above the line.
Now, we need to find the feasible region, which is the region that satisfies all the constraints. The feasible region is the overlapping region shaded in all three graphs.
Next, we need to find the corner points (vertices) of the feasible region. These corner points are the potential solutions to the linear programming problem.
To do this, we can find the intersection points of the different lines:
1. Intersection of lines 1 and 2: Solve the following system of equations:
x + y = 100
-x + y = 20
Adding these equations together, we get:
2y = 120
y = 60
Plugging this back into any of the equations, we find:
x + 60 = 100
x = 40
So, the intersection point is (40, 60).
2. Intersection of lines 1 and 3: Solve the following system of equations:
x + y = 100
-2x + 3y = 30
Multiplying the first equation by 2, we get:
2x + 2y = 200
Subtracting this equation from the second equation, we get:
5y = -170
y = -34
Plugging this back into any of the equations, we find:
x + (-34) = 100
x = 134
So, the intersection point is (134, -34).
The third intersection point, (0, 100), is already a corner point.
Now, we can calculate the value of the objective function g at each of the corner points to determine which one gives the minimum value:
1. For (40, 60):
g = 38(40) + 23(60) = 1520 + 1380 = 2900
2. For (134, -34):
g = 38(134) + 23(-34) = 5092 - 782 = 4310
3. For (0, 100):
g = 38(0) + 23(100) = 0 + 2300 = 2300
Therefore, the minimum value of g is 2300, and it occurs at the point (0, 100).
(x, y) = (0, 100)
g = 2300