2. What point in the feasible region maximizes the objective function?
Hint: graph constraints to find the region.
{y - 5 ≤ 4x
{y + x ≤ 10
{x ≥ 0, y ≥ 3
Maximum for
P = 7x - 5y
To find the point in the feasible region that maximizes the objective function P = 7x - 5y, we need to graph the constraints and find the intersection point(s) of all the inequalities.
First, let's graph each constraint:
1) y - 5 ≤ 4x
Rearranging the equation to slope-intercept form (y = mx + b), we have:
y ≤ 4x + 5
Plotting the line y = 4x + 5, we see that it has a y-intercept of (0,5) and a slope of 4.
2) y + x ≤ 10
Rearranging the equation to slope-intercept form, we have:
y ≤ -x + 10
Plotting the line y = -x + 10, we see that it has a y-intercept of (0,10) and a slope of -1.
3) x ≥ 0, y ≥ 3
These inequalities represent vertical and horizontal lines passing through (0,3).
Now, we need to find the intersection point(s) of all the inequalities. Looking at the graph, we can see that the feasible region is the shaded region where all three lines overlap.
However, since the objective function P = 7x - 5y is linear and not affected by the feasibility of the region, we can simply look for the highest value of P within the feasible region by evaluating P at the vertices of the feasible region.
The vertices of the feasible region are the points where the lines intersect. By analyzing the graph, we can find the following vertices:
- Intersection point of the two lines y = 4x + 5 and y = -x + 10: (1, 9)
- Intersection point of the line y = 4x + 5 with x = 0: (0, 5)
- Intersection point of the line y = -x + 10 with y = 3: (7, 3)
Next, we evaluate the objective function P = 7x - 5y at each vertex:
P(1, 9) = 7(1) - 5(9) = 7 - 45 = -38
P(0, 5) = 7(0) - 5(5) = 0 - 25 = -25
P(7, 3) = 7(7) - 5(3) = 49 - 15 = 34
Therefore, the point in the feasible region that maximizes the objective function P = 7x - 5y is (7, 3), with a value of P = 34.