By graphing the system of constraints, find the values of x and y that minimize the objective function.
x+2y>=8
x>=2
y>=0
minimum for C = x + 3y
To graph the system of constraints, we'll start with the inequalities:
1) x + 2y ≥ 8
2) x ≥ 2
3) y ≥ 0
First, let's graph the line for equality in the first constraint: x + 2y = 8.
To find the x and y-intercepts, let's set x = 0:
0 + 2y = 8
2y = 8
y = 4
So, one point on the line is (0, 4).
Now, let's set y = 0:
x + 2(0) = 8
x = 8
The second point on the line is (8, 0).
Drawing a line through these two points, we have the boundary line for the first constraint.
Next, let's graph the equality for the second constraint: x = 2.
This is a vertical line passing through (2,0).
Lastly, for the third constraint: y ≥ 0, this is the positive y-axis.
Graphing these lines and regions, we find:
|
|
| . O
|
--------------
| |
| |
|_____|
The shaded region represents the feasible region where all the constraints are satisfied.
Now, let's graph the objective function: C = x + 3y.
We can plot some points and connect them to make a line going through the feasible region as follows:
|
|
4 | . O
| |
3 | . | .
| /\
--------------
| |
|_____|
2 4
We can see that the line representing "C = x + 3y" is below the shaded region. The minimum value of the objective function occurs at the vertex in the shaded region closest to the line.
By analyzing, the vertex nearest to the line "C = x + 3y" is (2, 2).
Therefore, the values of x and y that minimize the objective function C = x + 3y are x = 2 and y = 2.
To find the values of x and y that minimize the objective function, we need to graph the system of constraints and identify the feasible region. Then, we can evaluate the objective function at each corner point of the feasible region to find the minimum value.
Let's graph the system of constraints:
1. Graph the line x + 2y = 8:
To graph this line, we can choose two points and draw a line passing through them.
When x = 0, we find y = 4, giving us the point (0, 4).
When y = 0, we find x = 8, giving us the point (8, 0).
2. Graph the line x = 2:
This line is vertical and passes through x = 2.
3. Graph the line y = 0:
This line is horizontal and passes through y = 0.
Now let's shade the feasible region:
- Since x + 2y ≥ 8, the feasible region is above or on the line x + 2y = 8.
- Since x ≥ 2, the feasible region is to the right or on the line x = 2.
- Since y ≥ 0, the feasible region is above or on the line y = 0.
The feasible region is the shaded area above or on the line x + 2y = 8, to the right or on the line x = 2, and above or on the line y = 0.
Now, we need to evaluate the objective function C = x + 3y at each corner point of the feasible region.
Let's find the intersection points of the lines:
1. Point A (0, 4):
C_A = 0 + 3(4) = 12
2. Point B (2, 3):
C_B = 2 + 3(3) = 11
3. Point C (8, 0):
C_C = 8 + 3(0) = 8
Thus, the minimum value of C = x + 3y is 8 when x = 8, y = 0.
To graph the system of constraints, we need to first rewrite the inequalities in slope-intercept form (y = mx + b).
1) x + 2y >= 8
Subtract x from both sides:
2y >= -x + 8
Divide both sides by 2:
y >= -0.5x + 4
2) x >= 2
This is already in the correct form.
3) y >= 0
This specifies that y should be greater than or equal to 0, which means it is a horizontal line at y = 0.
Now let's graph these three constraints.
1) Graph the line y = -0.5x + 4. To do this, plot the y-intercept (0, 4) and find another point. We can choose x = 2, which corresponds to y = 3. Connect these two points to draw the line.
2) Graph the line x = 2. Since it's a vertical line, draw a vertical line passing through x = 2.
3) Graph the line y = 0. This is the x-axis.
Now, we have a shaded region representing the feasible region. To find the values of x and y that minimize the objective function, we need to find the vertex of this region that gives the lowest value for C = x + 3y.
We can do this by locating the point where the lines intersect or by substituting the corner points of the feasible region into the objective function and finding the minimum value.
Let's find the vertex by calculating the values of (x, y) at each corner point.
The corner points of the feasible region are:
A: (2, 0)
B: (6, 0)
C: (2, 4)
Substituting these points into the objective function:
C(A) = 2 + 3(0) = 2
C(B) = 6 + 3(0) = 6
C(C) = 2 + 3(4) = 14
Therefore, the minimum value of C is 2, which occurs at point A (2, 0).
So, the values of x and y that minimize the objective function, C = x + 3y, are x = 2 and y = 0.