Solve the linear programming programing by graphing and then determining which vertex minimizes the objective function

C=3x+5y.

{x+y≥40
{2x+3y≥60
{x≥0
{y≥0

x =
y =

What is the minimum value? C=

input your data here for the solution, as well as a graph of the region:

http://www.zweigmedia.com/RealWorld/LPGrapher/lpg.html

To solve the linear programming problem, we will graph the given constraints and then determine the vertex that minimizes the objective function C = 3x + 5y.

Let's start by graphing the constraints:

1. x + y ≥ 40:
To graph this inequality, we will plot the line x + y = 40 and shade the region above it.

2. 2x + 3y ≥ 60:
To graph this inequality, we will plot the line 2x + 3y = 60 and shade the region above it.

3. x ≥ 0:
This implies that x should be greater than or equal to 0. Therefore, we draw a vertical line at x = 0.

4. y ≥ 0:
This implies that y should be greater than or equal to 0. Therefore, we draw a horizontal line at y = 0.

Now, let's find the vertices of the feasible region:

By visually examining the graph, we can see that the feasible region is the shaded region formed by the intersection of the two lines (x + y ≥ 40 and 2x + 3y ≥ 60) and the non-negative x and y axes.

The vertices of the feasible region are:
A(20, 20)
B(20, 40)
C(30, 30)

Next, we need to evaluate the objective function C = 3x + 5y at each of these vertices to find the minimum value.

C(A) = 3(20) + 5(20) = 60 + 100 = 160
C(B) = 3(20) + 5(40) = 60 + 200 = 260
C(C) = 3(30) + 5(30) = 90 + 150 = 240

Comparing the values, we can see that the minimum value of C occurs at vertex A(20, 20), where C = 160.

To solve the linear programming problem using graphing, we first need to graph the system of inequalities provided and then determine the feasible region. The feasible region is the region bounded by the inequalities.

To graph the inequalities, we can start by converting them to linear equations. Let's convert each inequality to an equation:

1. {x + y ≥ 40
Rewrite this equation in slope-intercept form (y = mx + b):
y ≥ -x + 40

2. {2x + 3y ≥ 60
Rewrite this equation in slope-intercept form (y = mx + b):
y ≥ (-2/3)x + 20

3. {x ≥ 0
This is a vertical line passing through the x-axis at x = 0.

4. {y ≥ 0
This is a horizontal line passing through the y-axis at y = 0.

Now, let's plot these equations on a graph:

Step 1: Plot the x-intercept and y-intercept of each line.

For the line y ≥ -x + 40, when x = 0, y = 40, so plot the point (0, 40). When y = 0, x = -40, so plot the point (-40, 0).

For the line y ≥ (-2/3)x + 20, when x = 0, y = 20, so plot the point (0, 20). When y = 0, x = 30, so plot the point (30, 0).

For the line x ≥ 0, plot a vertical line passing through x = 0.

For the line y ≥ 0, plot a horizontal line passing through y = 0.

Step 2: Shade the feasible region.

The feasible region is the overlapping region where all the inequalities hold true. Shade this region.

Step 3: Find the coordinates of the vertices.

The vertices are the points where the boundary lines intersect.

In this case, the feasible region is unbounded, so we cannot find the vertices.

However, to find the minimum value of the objective function C = 3x + 5y, we can evaluate the objective function at each corner of the feasible region and determine which one minimizes C.

Since we don't have vertices to evaluate, we cannot determine the minimum value C without solving the system algebraically instead of graphically.