Maximize `Z = 5x + 3y` subject to the constraints: `3x + 5y le 15, 5x + 2y le 10, x ge 0, y ge 0`

Answer

To solve the 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
The constraints given 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 boundary lines, we convert the inequalities into equations:
1. \( 3x + 5y = 15 \)
2. \( 5x + 2y = 10 \)

### Step 3: Find Intercepts
For each equation, we find the x-intercept and y-intercept.

**For \( 3x + 5y = 15 \):**
- Set \( y = 0 \): \( 3x = 15 \) → \( x = 5 \) (x-intercept: \( (5, 0) \))
- Set \( x = 0 \): \( 5y = 15 \) → \( y = 3 \) (y-intercept: \( (0, 3) \))

**For \( 5x + 2y = 10 \):**
- Set \( y = 0 \): \( 5x = 10 \) → \( x = 2 \) (x-intercept: \( (2, 0) \))
- Set \( x = 0 \): \( 2y = 10 \) → \( y = 5 \) (y-intercept: \( (0, 5) \))

### Step 4: Graph the Constraints
Plot the lines on a graph:
- The line \( 3x + 5y = 15 \) intersects at \( (5, 0) \) and \( (0, 3) \).
- The line \( 5x + 2y = 10 \) intersects at \( (2, 0) \) and \( (0, 5) \).

### Step 5: Determine the Feasible Region
The feasible region is where the inequalities overlap, bounded by the axes and the lines. This region will be in the first quadrant.

### Step 6: Find Intersection Points of the Lines
To find the intersection of the two lines, solve the equations simultaneously:
1. \( 3x + 5y = 15 \)
2. \( 5x + 2y = 10 \)

Multiply the first equation by 5 and the second by 3 to eliminate \( x \):
- \( 15x + 25y = 75 \)
- \( 15x + 6y = 30 \)

Subtract the second from the first:
\[ 19y = 45 \]
\[ y = \frac{45}{19} \]

Substituting \( y \) back into one of the original equations to find \( x \):
\[ 5x + 2\left(\frac{45}{19}\right) = 10 \]
\[ 5x + \frac{90}{19} = 10 \]
\[ 5x = 10 - \frac{90}{19} \]
\[ 5x = \frac{190 - 90}{19} = \frac{100}{19} \]
\[ x = \frac{20}{19} \]

Thus, the intersection point is \( \left(\frac{20}{19}, \frac{45}{19}\right) \).

### Step 7: Evaluate the Objective Function at Each Vertex
Evaluate \( Z = 5x + 3y \) at the vertices of the feasible region:
1. At \( (0, 3) \): \( Z = 5(0) + 3(3) = 9 \)
2. At \( (2, 0) \): \( Z = 5(2) + 3(0) = 10 \)
3. 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 8: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( \left(\frac{20}{19}, \frac{45}{19}\right) \) with \( Z = \frac{235}{19} \).

### Final Solution
The maximum value of \( Z \) is \( \frac{235}{19} \) at the point \( \left(\frac{20}{19}, \frac{45}{19}\right) \).

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