Maximum of `P=5x+7y` subject to the constraints :
`3x+2y le 12, 2x+3y le 13, x ge0, y ge0` is

Options

A. `91//3`
B. 31
C. 20
D. 120

Answer

To solve the linear programming problem of maximizing \( P = 5x + 7y \) subject to the constraints:

1. \( 3x + 2y \leq 12 \)
2. \( 2x + 3y \leq 13 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the Constraints
The constraints are:
- \( 3x + 2y \leq 12 \)
- \( 2x + 3y \leq 13 \)
- \( x \geq 0 \)
- \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
- \( 3x + 2y = 12 \)
- \( 2x + 3y = 13 \)

### Step 3: Find Intercepts of the Lines
For the first line \( 3x + 2y = 12 \):
- When \( x = 0 \): \( 2y = 12 \) → \( y = 6 \) (y-intercept)
- When \( y = 0 \): \( 3x = 12 \) → \( x = 4 \) (x-intercept)

So, the intercepts are \( (0, 6) \) and \( (4, 0) \).

For the second line \( 2x + 3y = 13 \):
- When \( x = 0 \): \( 3y = 13 \) → \( y = \frac{13}{3} \approx 4.33 \) (y-intercept)
- When \( y = 0 \): \( 2x = 13 \) → \( x = \frac{13}{2} = 6.5 \) (x-intercept)

So, the intercepts are \( (0, \frac{13}{3}) \) and \( (6.5, 0) \).

### Step 4: Plot the Lines and Identify the Feasible Region
Plot the lines on a graph and shade the feasible region that satisfies all constraints. The feasible region will be bounded by the axes and the lines.

### Step 5: Find the Corner Points of the Feasible Region
The corner points of the feasible region can be found as follows:
- Point A: \( (0, 0) \) (origin)
- Point B: \( (0, \frac{13}{3}) \) (intersection of y-axis with the second line)
- Point D: \( (4, 0) \) (intersection of x-axis with the first line)
- Point C: Intersection of the two lines \( 3x + 2y = 12 \) and \( 2x + 3y = 13 \).

To find point C, we can solve the equations simultaneously.

### Step 6: Solve for Intersection Point C
Multiply the first equation by 2 and the second by 3 to eliminate \( x \):
- \( 6x + 4y = 24 \) (from \( 3x + 2y = 12 \))
- \( 6x + 9y = 39 \) (from \( 2x + 3y = 13 \))

Subtract the first from the second:
\[ (6x + 9y) - (6x + 4y) = 39 - 24 \]
\[ 5y = 15 \implies y = 3 \]

Substituting \( y = 3 \) back into one of the original equations to find \( x \):
\[ 3x + 2(3) = 12 \implies 3x + 6 = 12 \implies 3x = 6 \implies x = 2 \]
So, point C is \( (2, 3) \).

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( P = 5x + 7y \) at each corner point:
- At A: \( P(0, 0) = 5(0) + 7(0) = 0 \)
- At B: \( P(0, \frac{13}{3}) = 5(0) + 7(\frac{13}{3}) = \frac{91}{3} \approx 30.33 \)
- At C: \( P(2, 3) = 5(2) + 7(3) = 10 + 21 = 31 \)
- At D: \( P(4, 0) = 5(4) + 7(0) = 20 \)

### Step 8: Determine the Maximum Value
The maximum value of \( P \) occurs at point C \( (2, 3) \) where \( P = 31 \).

### Conclusion
The maximum value of \( P = 5x + 7y \) subject to the given constraints is \( \boxed{31} \).