To find the maximum value of \( Z = 5x + 3y \) subject to the constraints \( 3x + 5y \leq 15 \), \( 5x + 2y \leq 10 \), and \( x, y \geq 0 \), we will follow these steps:
### Step 1: Identify the constraints
The constraints are:
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 intersection points, we convert the inequalities into equations:
1. \( 3x + 5y = 15 \)
2. \( 5x + 2y = 10 \)
### Step 3: Find the intercepts of the lines
For the first equation \( 3x + 5y = 15 \):
- When \( x = 0 \): \( 5y = 15 \) → \( y = 3 \) (Point: \( (0, 3) \))
- When \( y = 0 \): \( 3x = 15 \) → \( x = 5 \) (Point: \( (5, 0) \))
For the second equation \( 5x + 2y = 10 \):
- When \( x = 0 \): \( 2y = 10 \) → \( y = 5 \) (Point: \( (0, 5) \))
- When \( y = 0 \): \( 5x = 10 \) → \( x = 2 \) (Point: \( (2, 0) \))
### Step 4: Find the intersection of the two lines
To find the intersection point, we solve the system of equations:
1. \( 3x + 5y = 15 \)
2. \( 5x + 2y = 10 \)
We can multiply the first equation by 2 and the second equation by 5 to eliminate \( y \):
- \( 6x + 10y = 30 \) (from the first equation)
- \( 25x + 10y = 50 \) (from the second equation)
Now, subtract the first from the second:
\[
(25x + 10y) - (6x + 10y) = 50 - 30
\]
\[
19x = 20 \implies x = \frac{20}{19}
\]
Now substitute \( x = \frac{20}{19} \) back into one of the original equations to find \( y \):
Using \( 5x + 2y = 10 \):
\[
5\left(\frac{20}{19}\right) + 2y = 10
\]
\[
\frac{100}{19} + 2y = 10 \implies 2y = 10 - \frac{100}{19}
\]
\[
2y = \frac{190 - 100}{19} = \frac{90}{19} \implies y = \frac{45}{19}
\]
### Step 5: Identify the vertices of the feasible region
The vertices of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 3) \)
3. \( (2, 0) \)
4. \( \left(\frac{20}{19}, \frac{45}{19}\right) \)
### Step 6: Evaluate \( Z \) at each vertex
Now we calculate \( Z \) at each vertex:
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 \( \left(\frac{20}{19}, \frac{45}{19}\right) \):
\[
Z = 5\left(\frac{20}{19}\right) + 3\left(\frac{45}{19}\right) = \frac{100}{19} + \frac{135}{19} = \frac{235}{19} \approx 12.368
\]
### Step 7: Determine the maximum value
Comparing the values of \( Z \):
- \( Z(0, 0) = 0 \)
- \( Z(0, 3) = 9 \)
- \( Z(2, 0) = 10 \)
- \( Z\left(\frac{20}{19}, \frac{45}{19}\right) \approx 12.368 \)
The maximum value of \( Z \) is \( \frac{235}{19} \) or approximately \( 12.368 \).
### Final Answer:
The maximum value of \( Z \) is \( \frac{235}{19} \).
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