factory can produce two products, x and y, with a profit approximated by P=14x+22y−900 . The production of y must exceed the production of x by at least 100 units. Moreover, production levels are limited by the formula x+2y≤1400 .Identify the vertices of the feasible region.What production levels yield the maximum profit, and what is the maximum profit?
To find the vertices of the feasible region, we need to solve the system of inequalities:
x + 2y ≤ 1400
y ≥ x + 100
We can start by graphing these inequalities to get a visual representation of the feasible region.
First, let's graph the boundary line x + 2y = 1400:
To graph this line, we can start by finding its x-intercept and y-intercept.
Setting x = 0, we have:
0 + 2y = 1400
y = 700
So, one point on the line is (0, 700).
Setting y = 0, we have:
x + 2(0) = 1400
x = 1400
So, another point on the line is (1400, 0).
Plotting these two points and drawing the line, we have:
graph{(1400 - x)/2 [-500, 2000, -500, 2000]}
Next, let's graph the boundary line y = x + 100:
To graph this line, we can start by finding some points on the line.
Setting x = 0, we have:
y = 0 + 100
y = 100
So, one point on the line is (0, 100).
Setting x = 100, we have:
y = 100 + 100
y = 200
So, another point on the line is (100, 200).
Plotting these two points and drawing the line, we have:
graph{y = x + 100 [-500, 2000, -500, 2000]}
The feasible region is the shaded region where the two lines intersect and satisfy the given constraints:
graph{y = x + 100, (1400 - x)/2, y > x + 100 [-500, 2000, -500, 2000]}
Now, we can find the coordinates of the vertices of the feasible region by solving the equations of the boundary lines:
x + 2y = 1400
y = x + 100
Substituting y = x + 100 into x + 2y = 1400, we have:
x + 2(x + 100) = 1400
x + 2x + 200 = 1400
3x + 200 = 1400
3x = 1200
x = 400
Substituting x = 400 into y = x + 100, we have:
y = 400 + 100
y = 500
So, one vertex of the feasible region is (400, 500).
To find the other vertex, we can find the intersection of the two lines:
x + 2y = 1400 and y = x + 100
Substituting y = x + 100 into x + 2y = 1400, we have:
x + 2(x + 100) = 1400
x + 2x + 200 = 1400
3x + 200 = 1400
3x = 1200
x = 400
Substituting x = 400 into y = x + 100, we have:
y = 400 + 100
y = 500
So, the other vertex is also (400, 500).
Therefore, the vertices of the feasible region are (0, 700), (1400, 0), (400, 500), and (400, 500).
To find the production levels that yield the maximum profit, we need to evaluate the profit function P = 14x + 22y - 900 at each vertex.
At (0, 700):
P = 14(0) + 22(700) - 900
P = 15400
At (1400, 0):
P = 14(1400) + 22(0) - 900
P = 19600
At (400, 500):
P = 14(400) + 22(500) - 900
P = 14100
At (400, 500):
P = 14(400) + 22(500) - 900
P = 14100
Therefore, the production levels that yield the maximum profit are x = 1400 and y = 0, and the maximum profit is $19,600.