The maximum value of `2x+y` subject to
`3x+5yle26and5x+3yle30,xge0,yge0` is

Options

A. 12
B. 11.5
C. 10
D. 17.33

Explanation

To find the maximum value of \(2x + y\) subject to the constraints given, we will follow these steps:

### Step 1: Write down the constraints
The constraints are:
1. \(3x + 5y \leq 26\)
2. \(5x + 3y \leq 30\)
3. \(x \geq 0\)
4. \(y \geq 0\)

### Step 2: Convert inequalities to equations
To find the boundary lines of the inequalities, we convert them into equations:
1. \(3x + 5y = 26\)
2. \(5x + 3y = 30\)

### Step 3: Find the intercepts of the lines
For the first equation \(3x + 5y = 26\):
- When \(x = 0\), \(5y = 26 \Rightarrow y = \frac{26}{5} = 5.2\)
- When \(y = 0\), \(3x = 26 \Rightarrow x = \frac{26}{3} \approx 8.67\)

For the second equation \(5x + 3y = 30\):
- When \(x = 0\), \(3y = 30 \Rightarrow y = 10\)
- When \(y = 0\), \(5x = 30 \Rightarrow x = 6\)

### Step 4: Plot the lines and find the feasible region
Now we can plot the lines on a graph:
- The line \(3x + 5y = 26\) intersects the axes at \((0, 5.2)\) and \((8.67, 0)\).
- The line \(5x + 3y = 30\) intersects the axes at \((0, 10)\) and \((6, 0)\).

Next, we need to determine the feasible region that satisfies all the inequalities. This is the area where both inequalities overlap, and it will be bounded by the axes and the lines.

### Step 5: Find the intersection point of the two lines
To find the intersection point of the two lines, we solve the equations simultaneously:
1. \(3x + 5y = 26\)
2. \(5x + 3y = 30\)

We can solve this system using substitution or elimination. Let's use elimination:
- Multiply the first equation by 3: \(9x + 15y = 78\)
- Multiply the second equation by 5: \(25x + 15y = 150\)

Now, subtract the first from the second:
\[ (25x + 15y) - (9x + 15y) = 150 - 78 \]
\[ 16x = 72 \Rightarrow x = \frac{72}{16} = 4.5 \]

Substituting \(x = 4.5\) back into one of the original equations to find \(y\):
\[ 3(4.5) + 5y = 26 \Rightarrow 13.5 + 5y = 26 \Rightarrow 5y = 12.5 \Rightarrow y = 2.5 \]
Thus, the intersection point is \((4.5, 2.5)\).

### Step 6: Evaluate the objective function at the corner points
Now we need to evaluate \(2x + y\) at the corner points of the feasible region:
1. \((0, 0)\): \(2(0) + 0 = 0\)
2. \((6, 0)\): \(2(6) + 0 = 12\)
3. \((0, 5.2)\): \(2(0) + 5.2 = 5.2\)
4. \((4.5, 2.5)\): \(2(4.5) + 2.5 = 9 + 2.5 = 11.5\)

### Step 7: Determine the maximum value
Comparing all the values:
- At \((0, 0)\) → \(0\)
- At \((6, 0)\) → \(12\)
- At \((0, 5.2)\) → \(5.2\)
- At \((4.5, 2.5)\) → \(11.5\)

The maximum value of \(2x + y\) is \(12\) at the point \((6, 0)\).

### Final Answer
The maximum value of \(2x + y\) subject to the given constraints is **12**.

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