What point maximizes the objective function

P = 4x + 3y for the constraints below. What is the maximum value?

Constraints:

x>=0
y>=0
y<=1/3x-4
6>=y

To find the point that maximizes the objective function P = 4x + 3y, we need to graph the constraints and identify their feasible region.

1. Graph the constraint y <= 1/3x - 4:
- Begin by plotting the line y = 1/3x - 4.
- Choose an x-intercept; let's say x = 6.
- When x = 6, y = 1/3(6) - 4 = -2.
- Plot the points (6, -2) and (0, -4) and draw a line through them.
- Shade the area below the line, as it represents the constraint y <= 1/3x - 4.

2. Graph the constraint y <= 6:
- Draw a horizontal line at y = 6.
- Shade the area below the line as it represents the constraint y <= 6.

3. Graph the constraint x >= 0 and y >= 0:
- Draw the x-axis (x >= 0) and the y-axis (y >= 0).
- Shade the entire positive quadrant, as it represents the constraints x >= 0 and y >= 0.

4. Identify the feasible region:
- The feasible region is the intersection of the shaded areas from the previous steps. It is the area that satisfies all the given constraints.

5. Find the coordinate(s) of the corner point(s) within the feasible region:
- Locate the corners of the feasible region (vertices).
- For this case, the corners are where the lines intersect or touch the axes.

6. Calculate the objective function at each corner point:
- Evaluate the objective function P = 4x + 3y at each corner point.
- Determine which corner point gives the maximum value for P.

7. Identify the maximum value:
- The maximum value of P is the largest value obtained from step 6.

By following these steps, you should be able to find the point that maximizes the objective function P = 4x + 3y and determine its maximum value.

To find the point that maximizes the objective function P = 4x + 3y, we need to analyze the given constraints and graph them to determine the feasible region. The maximum value of P can then be determined by evaluating the objective function at each corner point of the feasible region.

1. The first constraint, x >= 0, indicates that x must be greater than or equal to 0. This implies that x will be along or to the right of the y-axis.

2. The second constraint, y >= 0, implies that y must be greater than or equal to 0. This means that y will be either on or above the x-axis.

3. The third constraint, y <= (1/3)x - 4, represents a line in slope-intercept form (y = mx + b). To graph this constraint, we can start by plotting the y-intercept, which is -4. Then, using the slope of 1/3, we can determine additional points to plot the line. Since the constraint is inclusive (y <= ...), the line will be below or on this line.

4. The fourth constraint is simply 6 >= y, which means that y must be less than or equal to 6. This constraint indicates that the feasible region will be vertically restricted to below or on the line y = 6.

Now, we can graph these constraints on a coordinate plane and find the intersection points.

To find the intersection points:

1. Locate the points where the lines y = (1/3)x - 4 and y = 6 intersect. To do this, set the two equations equal to each other:
(1/3)x - 4 = 6
(1/3)x = 10
x = 30

Substituting x = 30 into either of the equations, we get y = (1/3)(30) - 4 = 6.

Therefore, the intersection point is (30, 6).

2. We can also find the x-intercept of the line y = (1/3)x - 4 by setting y = 0 and solving for x:
0 = (1/3)x - 4
(1/3)x = 4
x = 12

Substituting x = 12 into the equation, we get y = (1/3)(12) - 4 = 0.

Therefore, the intersection point is (12, 0).

Now, we have three points: (0, 0), (12, 0), and (30, 6). These represent the corners of the feasible region.

Finally, we can evaluate the objective function P = 4x + 3y at each corner point to find the maximum value.

- P at (0, 0): P = 4(0) + 3(0) = 0
- P at (12, 0): P = 4(12) + 3(0) = 48
- P at (30, 6): P = 4(30) + 3(6) = 132

Therefore, the maximum value of P is 132, which occurs at the point (30, 6).

Graph the inequalities given by the set of constraints. Find points where the boundary lines intersect to form a polygon. Substitute the coordinates of each point into the objective function and find the one that results in the largest value.