Answer: To find the minimum value of the equation Q = 5x + 6y + 1.5, subject to the given constraints, we need to analyze the feasible region formed by the intersection of the three inequalities.
The constraints are:
x ≥ 4
-3x + 4y ≥ -8
3x + 4y ≤ 16
Let's first graph these inequalities to visualize the feasible region:
Graph x ≥ 4:
This is a vertical line passing through x = 4, including all points to the right of it.
Graph -3x + 4y ≥ -8:
To graph this inequality, let's rewrite it as 4y ≥ 3x - 8, and then y ≥ (3/4)x - 2. This is the equation of a line with a slope of 3/4 and y-intercept of -2. Shade the region above the line.
Graph 3x + 4y ≤ 16:
To graph this inequality, let's rewrite it as 4y ≤ -3x + 16, and then y ≤ (-3/4)x + 4. This is the equation of a line with a slope of -3/4 and y-intercept of 4. Shade the region below the line.
Now, we can identify the feasible region, which is the area shaded by all three inequalities.
The minimum value of Q = 5x + 6y + 1.5 occurs at one of the corner points of the feasible region. Let's find the coordinates of these corner points by solving the system of equations formed by the lines where the inequalities intersect.
Intersection of x = 4 and y ≥ (3/4)x - 2:
Substitute x = 4 into y = (3/4)x - 2:
y = (3/4) * 4 - 2
y = 3 - 2
y = 1
Coordinate: (4, 1)
Intersection of x = 4 and y ≤ (-3/4)x + 4:
Substitute x = 4 into y = (-3/4)x + 4:
y = (-3/4) * 4 + 4
y = -3 + 4
y = 1
Coordinate: (4, 1)
Intersection of -3x + 4y ≥ -8 and 3x + 4y ≤ 16:
Solve the system of equations:
-3x + 4y = -8
3x + 4y = 16
Add the equations to eliminate x:
(4y = 8) -> y = 2
Substitute y = 2 into either equation to find x:
-3x + 4(2) = -8
-3x + 8 = -8
-3x = -16
x = 16/3
Coordinate: (16/3, 2)
Now, we need to evaluate Q = 5x + 6y + 1.5 at each of these coordinates to find the minimum value:
Q at (4, 1):
Q = 5(4) + 6(1) + 1.5
Q = 20 + 6 + 1.5
Q = 27.5
Q at (16/3, 2):
Q = 5(16/3) + 6(2) + 1.5
Q = 80/3 + 12 + 1.5
Q = 25.5
The minimum value of Q occurs at the point (16/3, 2) with a value of approximately 25.5.