HELP? Exam in 20 mins -dont' have any idea how to do this question; please helpp.. anyone!
PLEAZE A.S.A.P.
minimize p= 15x +18y subject to x+2y<=20 3x + 2y>= 36 x>=0 y>=0
To solve the given problem and minimize the objective function p = 15x + 18y, subject to the constraints x + 2y ≤ 20, 3x + 2y ≥ 36, x ≥ 0, and y ≥ 0, you can follow these steps:
Step 1: Graph the constraints
Graph each constraint on a graph paper to visualize the feasible region. Start with the inequality x + 2y ≤ 20:
First, rearrange the inequality to solve for y:
2y ≤ 20 − x
y ≤ (20 − x)/2
Now, plot the boundary line x + 2y = 20, which is obtained by replacing the inequality sign with an equal sign:
x + 2y = 20
y = (20 − x)/2
Next, shade the region below the line, including any points on the line that satisfy the inequality.
Next, graph the inequality 3x + 2y ≥ 36:
Rearrange the inequality to solve for y:
2y ≥ 36 − 3x
y ≥ (36 − 3x)/2
Now, plot the boundary line 3x + 2y = 36, which is obtained by replacing the inequality sign with an equal sign:
3x + 2y = 36
y = (36 − 3x)/2
Finally, shade the region above the line, including any points on the line that satisfy the inequality.
Step 2: Identify the feasible region
The feasible region is the overlapping shaded region formed by the intersection of the two shaded areas from Step 1.
Step 3: Identify the corner points of the feasible region
The corner points of the feasible region are the vertices where the boundary lines intersect. In this case, we have two corner points, (0, 10) and (8, 4).
Step 4: Evaluate the objective function at each corner point
Now, substitute the corner points into the objective function p = 15x + 18y to find the minimum value of p.
For the first corner point, (0, 10):
p = 15(0) + 18(10)
p = 0 + 180
p = 180
For the second corner point, (8, 4):
p = 15(8) + 18(4)
p = 120 + 72
p = 192
Therefore, the minimum value of p is 180, and it occurs at the corner point (0, 10).
I hope this explanation helps you solve the problem in your exam. Good luck!