To solve the linear programming problem of maximizing \( Z = 5x + 3y \) subject to the constraints \( 3x + 5y \leq 15 \), \( 5x + 2y \leq 10 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:
### Step 1: Identify the Constraints
We have the following constraints:
1. \( 3x + 5y \leq 15 \)
2. \( 5x + 2y \leq 10 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
### Step 2: Convert Inequalities to Equations
To find the boundary lines of the inequalities, we convert the inequalities into equations:
1. \( 3x + 5y = 15 \)
2. \( 5x + 2y = 10 \)
### Step 3: Find Intercepts
For each equation, we will find the x-intercept and y-intercept.
**For \( 3x + 5y = 15 \):**
- When \( x = 0 \): \( 5y = 15 \) → \( y = 3 \) (Point: \( (0, 3) \))
- When \( y = 0 \): \( 3x = 15 \) → \( x = 5 \) (Point: \( (5, 0) \))
**For \( 5x + 2y = 10 \):**
- When \( x = 0 \): \( 2y = 10 \) → \( y = 5 \) (Point: \( (0, 5) \))
- When \( y = 0 \): \( 5x = 10 \) → \( x = 2 \) (Point: \( (2, 0) \))
### Step 4: Plot the Lines
Plot the points \( (0, 3) \), \( (5, 0) \), \( (0, 5) \), and \( (2, 0) \) on a graph and draw the lines for the equations.
### Step 5: Determine the Feasible Region
To find the feasible region, we need to check which side of the lines satisfies the inequalities. We can test the origin \( (0, 0) \):
- For \( 3x + 5y \leq 15 \): \( 3(0) + 5(0) = 0 \leq 15 \) (True)
- For \( 5x + 2y \leq 10 \): \( 5(0) + 2(0) = 0 \leq 10 \) (True)
Since the origin satisfies both inequalities, the feasible region is below both lines and in the first quadrant.
### Step 6: Find Corner Points
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 3) \)
3. \( (2, 0) \)
4. The intersection of the lines \( (3x + 5y = 15) \) and \( (5x + 2y = 10) \).
### Step 7: Solve for Intersection Point
To find the intersection point, we solve the equations:
1. \( 3x + 5y = 15 \)
2. \( 5x + 2y = 10 \)
We can multiply the first equation by 2 and the second by 5:
- \( 6x + 10y = 30 \)
- \( 25x + 10y = 50 \)
Now, subtract the first from the second:
\[
25x + 10y - (6x + 10y) = 50 - 30
\]
\[
19x = 20 \implies x = \frac{20}{19}
\]
Substituting \( x \) back into one of the equations to find \( y \):
\[
3\left(\frac{20}{19}\right) + 5y = 15
\]
\[
\frac{60}{19} + 5y = 15
\]
\[
5y = 15 - \frac{60}{19} = \frac{285 - 60}{19} = \frac{225}{19} \implies y = \frac{45}{19}
\]
Thus, the intersection point is \( \left(\frac{20}{19}, \frac{45}{19}\right) \).
### Step 8: Evaluate Z at Each Corner Point
Now we evaluate \( Z = 5x + 3y \) at each corner point:
1. At \( (0, 0) \): \( Z = 5(0) + 3(0) = 0 \)
2. At \( (0, 3) \): \( Z = 5(0) + 3(3) = 9 \)
3. At \( (2, 0) \): \( Z = 5(2) + 3(0) = 10 \)
4. At \( \left(\frac{20}{19}, \frac{45}{19}\right) \):
\[
Z = 5\left(\frac{20}{19}\right) + 3\left(\frac{45}{19}\right) = \frac{100 + 135}{19} = \frac{235}{19} \approx 12.368
\]
### Step 9: Determine Maximum Z
The maximum value of \( Z \) occurs at \( \left(\frac{20}{19}, \frac{45}{19}\right) \) with \( Z \approx 12.368 \).
### Final Answer
The maximum value of \( Z \) is \( \frac{235}{19} \) or approximately \( 12.368 \) at the point \( \left(\frac{20}{19}, \frac{45}{19}\right) \).
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