An objective function and a system of linear inequalities representing constraints are given. Graph the system of inequalities representing the constraints. Find the value of the objective function at each corner of the graphed region. Use these values to determine the maximum value of the objective function and the values of x and y for which the maximum occurs.

Objective Function z = 5x - 23y
Constraints 0 ¡Ü x ¡Ü 5
0 ¡Ü y ¡Ü 8
4x + 5y ¡Ü 30
4x + 3y ¡Ü 20

I do not understand your symbols for the constraint functions.

Suspect you are doing a linear programming problem.
I will look up a reference for how to do linear programming.

Perhaps this will help:

http://people.hofstra.edu/Stefan_Waner/RealWorld/Summary4.html

To graph the system of inequalities, start by graphing the four constraint inequalities one by one.

First, graph the inequality 0 <= x <= 5 by drawing a horizontal line segment from (0,0) to (5,0).

Next, graph the inequality 0 <= y <= 8 by drawing a vertical line segment from (0,0) to (0,8).

Now, graph the inequality 4x + 5y <= 30 by rearranging it to get y <= (-4/5)x + 6. To graph this line, find the x- and y-intercepts by setting x and y to 0 respectively: (-1.5, 0) and (0, 6). Connect these points with a line and shade below it to represent the region y <= (-4/5)x + 6.

Finally, graph the inequality 4x + 3y <= 20 by rearranging it to get y <= (-4/3)x + 20/3. Find the x- and y-intercepts: (-5, 0) and (0, 20/3). Connect these points with a line and shade below it to represent the region y <= (-4/3)x + 20/3.

By graphing the four inequalities, you will have four lines and a shaded region bounded by them.

Now, find the corners of the shaded region:

- (0, 0)
- (0, 8)
- (5, 0)
- (3, 4) (the intersection point of the two lines)

Plug the x and y values of each corner into the objective function z = 5x - 23y:

At (0, 0): z = 5(0) - 23(0) = 0
At (0, 8): z = 5(0) - 23(8) = -184
At (5, 0): z = 5(5) - 23(0) = 25
At (3, 4): z = 5(3) - 23(4) = -67

From these values, we can see that the maximum value of the objective function z = 5x - 23y occurs at (5, 0) with a value of 25.

To graph the system of linear inequalities, we will plot the lines representing the constraints and shade the region that satisfies all the inequalities.

First, let's graph the constraint 0 ≤ x ≤ 5:
- Draw a vertical line passing through x = 0 (the y-axis) and another vertical line passing through x = 5. Shade the region between these two lines.

Next, let's graph the constraint 0 ≤ y ≤ 8:
- Draw a horizontal line passing through y = 0 (the x-axis) and another horizontal line passing through y = 8. Shade the region between these two lines.

The next two constraints involve lines, so we need to find their equations.

For the constraint 4x + 5y ≤ 30:
- Let's find the equation of the line by rearranging it into slope-intercept form (y = mx + b):
4x + 5y ≤ 30
5y ≤ -4x + 30
y ≤ (-4/5)x + 6

For the constraint 4x + 3y ≤ 20:
- Let's find the equation of the line by rearranging it into slope-intercept form (y = mx + b):
4x + 3y ≤ 20
3y ≤ -4x + 20
y ≤ (-4/3)x + 20/3

Now, plot the lines y = (-4/5)x + 6 and y = (-4/3)x + 20/3. Shade the region below these lines, as we are looking for values that satisfy the inequalities.

The shaded region should look something like a trapezoid.

Next, we need to find the corners of the graphed region. These corners are the points where the lines intersect.

Label the corners A, B, C, D, E:
A: The intersection of the lines x = 0 and y = 0 (bottom left corner).
B: The intersection of the lines x = 0 and y = 8 (top left corner).
C: The intersection of the lines x = 5 and y = 0 (bottom right corner).
D: The intersection of the lines y = (-4/5)x + 6 and y = 8 (top right corner).
E: The intersection of the lines y = (-4/3)x + 20/3 and y = 8 (top right corner).

Now, let's find the value of the objective function z = 5x - 23y at each of these corner points.

For point A (0, 0):
z = 5(0) - 23(0) = 0

For point B (0, 8):
z = 5(0) - 23(8) = -184

For point C (5, 0):
z = 5(5) - 23(0) = 25

For point D:
First, solve the system of equations (-4/5)x + 6 = 8 and y = (-4/5)x + 6 simultaneously.
(-4/5)x + 6 = 8
-4x + 30 = 40
-4x = 10
x = -2.5
y = (-4/5)(-2.5) + 6 = 8

z = 5(-2.5) - 23(8) = -185

For point E:
First, solve the system of equations (-4/3)x + 20/3 = 8 and y = (-4/3)x + 20/3 simultaneously.
(-4/3)x + 20/3 = 8
-4x + 20 = 24
-4x = 4
x = -1
y = (-4/3)(-1) + 20/3 = 8

z = 5(-1) - 23(8) = -184

Now, we compare the values of the objective function at each corner point.

The maximum value of the objective function is z = 25 at point C (5, 0). Therefore, the maximum value of z is 25, and the values of x and y for which the maximum occurs are x = 5 and y = 0.