If `3x_(1)+5x_(2) le 15,5x_(1)+2x_(2) le 10,x_(1),x_(2)ge0,` then the maximum value of `z=5x_(1)+3x_(2)` by graphical method is

Options

A. `12(7)/(19)`
B. `12(1)/(7)`
C. `12(3)/(5)`
D. `12`

Explanation

To solve the problem using the graphical method, we need to follow these steps:

### Step 1: Write down the inequalities
We have the following inequalities:
1. \(3x_1 + 5x_2 \leq 15\)
2. \(5x_1 + 2x_2 \leq 10\)
3. \(x_1, x_2 \geq 0\)

### Step 2: Convert inequalities to equations
To graph these inequalities, we first convert them to equations:
1. \(3x_1 + 5x_2 = 15\)
2. \(5x_1 + 2x_2 = 10\)

### Step 3: Find intercepts for the first equation
For the equation \(3x_1 + 5x_2 = 15\):
- When \(x_1 = 0\):
\[ 5x_2 = 15 \implies x_2 = 3 \quad \text{(Point: (0, 3))} \]
- When \(x_2 = 0\):
\[ 3x_1 = 15 \implies x_1 = 5 \quad \text{(Point: (5, 0))} \]

### Step 4: Find intercepts for the second equation
For the equation \(5x_1 + 2x_2 = 10\):
- When \(x_1 = 0\):
\[ 2x_2 = 10 \implies x_2 = 5 \quad \text{(Point: (0, 5))} \]
- When \(x_2 = 0\):
\[ 5x_1 = 10 \implies x_1 = 2 \quad \text{(Point: (2, 0))} \]

### Step 5: Plot the lines on a graph
Now, we plot the points:
- For \(3x_1 + 5x_2 = 15\), plot (0, 3) and (5, 0).
- For \(5x_1 + 2x_2 = 10\), plot (0, 5) and (2, 0).

### Step 6: Determine the feasible region
The feasible region is where the shaded areas of both inequalities overlap, bounded by the axes.

### Step 7: Find the intersection point of the lines
To find the intersection of the two lines, we can solve the equations:
1. \(3x_1 + 5x_2 = 15\)
2. \(5x_1 + 2x_2 = 10\)

We can multiply the first equation by 5 and the second by 3 to eliminate \(x_1\):
\[ 15x_1 + 25x_2 = 75 \quad \text{(1)} \]
\[ 15x_1 + 6x_2 = 30 \quad \text{(2)} \]

Subtract equation (2) from (1):
\[ (25x_2 - 6x_2) = 75 - 30 \implies 19x_2 = 45 \implies x_2 = \frac{45}{19} \]

Substituting \(x_2\) back into the first equation to find \(x_1\):
\[ 3x_1 + 5\left(\frac{45}{19}\right) = 15 \]
\[ 3x_1 + \frac{225}{19} = 15 \]
\[ 3x_1 = 15 - \frac{225}{19} = \frac{285 - 225}{19} = \frac{60}{19} \]
\[ x_1 = \frac{20}{19} \]

### Step 8: Evaluate \(z = 5x_1 + 3x_2\)
Now, we evaluate \(z\) at the vertices of the feasible region:
1. At (0, 3):
\[ z = 5(0) + 3(3) = 9 \]
2. At (5, 0):
\[ z = 5(5) + 3(0) = 25 \]
3. At (0, 5):
\[ z = 5(0) + 3(5) = 15 \]
4. At (2, 0):
\[ z = 5(2) + 3(0) = 10 \]
5. 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 the maximum value
The maximum value of \(z\) occurs at the point \(\left(\frac{20}{19}, \frac{45}{19}\right)\) with:
\[ z = \frac{235}{19} \]

### Conclusion
The maximum value of \(z\) is \(\frac{235}{19}\).