Graph the system of constraints and find the value of x and y that maximize the objective function
x ≥ 0
y ≥ 0
y ≤ 3
y ≤ -x+5
objective function: C=-6x+2y
To graph the system of constraints, we first need to plot the lines representing the constraints on a coordinate plane.
First constraint: x ≥ 0
This constraint indicates that the x-values must be greater than or equal to zero. This is the x-axis.
Second constraint: y ≥ 0
This constraint indicates that the y-values must be greater than or equal to zero. This is the y-axis.
Third constraint: y ≤ 3
This constraint indicates that the y-values must be less than or equal to 3. This is a horizontal line passing through the point (0, 3).
Fourth constraint: y ≤ -x + 5
This constraint indicates that the y-values must be less than or equal to the negative slope of -1 and the y-intercept of 5. This line passes through the points (5, 0) and (0, 5).
Now, let's graph the system of constraints:
Here is the graph:
(https://latex.codecogs.com/svg.latex?%5Cbegin%7Btikzpicture%7D%5Bscale%3D0.8%5D%0A%20%20%5Cdraw%20%5B-%3E%2C%20thick%5D%20(-1%2C0)%20--%20(6%2C0)%20node%20%5Bbelow%5D%20%7Bx%7D%3B%0A%20%20%5Cdraw%20%5B-%3E%2C%20thick%5D%20(0%2C-1)%20--%20(0%2C6)%20node%20%5Bleft%5D%20%7By%7D%3B%0A%20%20%5Cdraw%20%5Bthick%2C%20dashed%5D%20(3%2C0)%20--%20(3%2C5)%3B%0A%20%20%5Cdraw%20%5Bthick%2C%20dashed%5D%20(0%2C3)%20--%20(5%2C0)%3B%0A%20%20%5Cdraw%20%5B%2C%20thick%5D%20(0%2C3)%20node%20%5Babove%5D%20%7By+%3D%203%7D%3B%0A%20%20%5Cdraw%20%5B%2C%20thick%5D%20(4.3%2C2.3)%20node%20%5Babove%5D%20%7By%20%3D%20-x%20%2B%205%7D%3B%0A%20%20%5Cforeach%20%5Cx%20in%20%7B0%2C1%2C2%2C3%2C4%2C5%7D%0A%20%20%20%20%5Cdraw%20(%5Cx%2C-0.2)%20--%20(%5Cx%2C0.2)%3B%0A%20%20%5Cforeach%20%5Cy%20in%20%7B0%2C1%2C2%2C3%2C4%2C5%7D%0A%20%20%20%20%5Cdraw%20(-0.2%2C%5Cy)%20--%20(0.2%2C%5Cy)%3B%0A%5Cend%7Btikzpicture%7D)
To find the value of x and y that maximize the objective function, we need to locate the feasible region, which is the area that satisfies all the constraints. In this case, the feasible region is the triangle formed by the lines y = 3, y = -x + 5, and the x-axis.
Next, we substitute the vertices of the feasible region in the objective function to determine which values of x and y maximize the objective function.
To find the vertices of the feasible region, we solve the system of equations formed by the lines' intersections.
First, we find the intersection of y = 3 and y = -x + 5:
y = 3
- x + 5 = 3
- x = 3 - 5
x = 2
Therefore, the vertex (2, 3) is one of the points intersecting the lines.
Next, we find the intersection of y = 3 and the x-axis:
y = 3
Therefore, the vertex (0, 3) is the other point intersecting the lines.
Finally, we find the intersection of y = -x + 5 and the x-axis:
y = 0
- x + 5 = 0
- x = -5
x = 5
Therefore, the vertex (5, 0) is the last point intersecting the lines.
Substituting these vertices into the objective function C = -6x + 2y:
C(2, 3) = -6(2) + 2(3) = -12 + 6 = -6
C(0, 3) = -6(0) + 2(3) = 6
C(5, 0) = -6(5) + 2(0) = -30
Thus, the maximum value of the objective function is 6, and it occurs at the point (0, 3).