To solve the given linear programming problem graphically, we will follow these steps:
### Step 1: Formulate the Problem
We need to minimize the objective function:
\[ z = 3x + 5y \]
Subject to the constraints:
1. \( x + y \geq 2 \)
2. \( x + 3y \geq 3 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
### Step 2: Convert Inequalities to Equations
Convert the inequalities into equations to find the boundary lines:
1. \( x + y = 2 \)
2. \( x + 3y = 3 \)
### Step 3: Find Intercepts for Each Line
**For the line \( x + y = 2 \)**:
- Set \( x = 0 \):
\[ 0 + y = 2 \Rightarrow y = 2 \]
Point: \( (0, 2) \)
- Set \( y = 0 \):
\[ x + 0 = 2 \Rightarrow x = 2 \]
Point: \( (2, 0) \)
**For the line \( x + 3y = 3 \)**:
- Set \( x = 0 \):
\[ 0 + 3y = 3 \Rightarrow y = 1 \]
Point: \( (0, 1) \)
- Set \( y = 0 \):
\[ x + 0 = 3 \Rightarrow x = 3 \]
Point: \( (3, 0) \)
### Step 4: Plot the Lines on a Graph
Plot the points \( (0, 2) \), \( (2, 0) \), \( (0, 1) \), and \( (3, 0) \) on a graph. Draw the lines for the equations \( x + y = 2 \) and \( x + 3y = 3 \).
### Step 5: Determine the Feasible Region
To find the feasible region, we will test a point (e.g., \( (0, 0) \)):
- For \( x + y \geq 2 \):
\[ 0 + 0 \geq 2 \quad \text{(False)} \]
- For \( x + 3y \geq 3 \):
\[ 0 + 0 \geq 3 \quad \text{(False)} \]
Since both conditions are false, the feasible region is away from the origin. Shade the region that satisfies both inequalities.
### Step 6: Identify Corner Points
The corner points of the feasible region are:
1. \( (0, 2) \)
2. \( (2, 0) \)
3. Intersection of \( x + y = 2 \) and \( x + 3y = 3 \)
### Step 7: Find the Intersection Point
To find the intersection of the lines \( x + y = 2 \) and \( x + 3y = 3 \):
1. Solve the system of equations:
\[
x + y = 2 \quad (1)
\]
\[
x + 3y = 3 \quad (2)
\]
Subtract (1) from (2):
\[
(x + 3y) - (x + y) = 3 - 2 \Rightarrow 2y = 1 \Rightarrow y = \frac{1}{2}
\]
Substitute \( y = \frac{1}{2} \) into (1):
\[
x + \frac{1}{2} = 2 \Rightarrow x = \frac{3}{2}
\]
Intersection Point: \( \left(\frac{3}{2}, \frac{1}{2}\right) \)
### Step 8: Evaluate the Objective Function at Each Corner Point
1. At \( (0, 2) \):
\[ z = 3(0) + 5(2) = 10 \]
2. At \( (2, 0) \):
\[ z = 3(2) + 5(0) = 6 \]
3. At \( \left(\frac{3}{2}, \frac{1}{2}\right) \):
\[ z = 3\left(\frac{3}{2}\right) + 5\left(\frac{1}{2}\right) = \frac{9}{2} + \frac{5}{2} = 7 \]
### Step 9: Determine the Minimum Value
The minimum value of \( z \) occurs at the point \( \left(\frac{3}{2}, \frac{1}{2}\right) \) with:
\[ \text{Minimum } z = 7 \]
### Final Result
The minimum value of the objective function \( z = 3x + 5y \) is \( 7 \) at the point \( \left(\frac{3}{2}, \frac{1}{2}\right) \).
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