Maximize `Z= 4x + 45y` subject to :
`x + y le 300, 2x + 3y le 7, x, y ge 0`.

Answer

To solve the problem of maximizing \( Z = 4x + 45y \) subject to the constraints:

1. \( x + y \leq 300 \)
2. \( 2x + 3y \leq 7 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the Constraints
We have the following constraints:
- \( x + y \leq 300 \)
- \( 2x + 3y \leq 7 \)
- \( x \geq 0 \)
- \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( x + y = 300 \)
2. \( 2x + 3y = 7 \)

### Step 3: Find Intercepts
For the first equation \( x + y = 300 \):
- When \( x = 0 \), \( y = 300 \) (Point A: \( (0, 300) \))
- When \( y = 0 \), \( x = 300 \) (Point B: \( (300, 0) \))

For the second equation \( 2x + 3y = 7 \):
- When \( x = 0 \), \( y = \frac{7}{3} \approx 2.33 \) (Point C: \( (0, \frac{7}{3}) \))
- When \( y = 0 \), \( x = \frac{7}{2} = 3.5 \) (Point D: \( (\frac{7}{2}, 0) \))

### Step 4: Graph the Constraints
Plot the points on a graph:
- Draw the line for \( x + y = 300 \) through points A and B.
- Draw the line for \( 2x + 3y = 7 \) through points C and D.

### Step 5: Identify the Feasible Region
The feasible region is where all constraints overlap, which will be in the first quadrant (since \( x \geq 0 \) and \( y \geq 0 \)).

### Step 6: Find the Corner Points of the Feasible Region
The corner points of the feasible region can be found by solving the equations:
1. Intersection of \( x + y = 300 \) and \( 2x + 3y = 7 \).

To find the intersection:
- From \( x + y = 300 \), we can express \( y = 300 - x \).
- Substitute into \( 2x + 3(300 - x) = 7 \):
\[ 2x + 900 - 3x = 7 \\ -x + 900 = 7 \\ x = 893 \]
This value is not valid since it exceeds the constraint \( x + y \leq 300 \).

Next, check the corner points:
1. \( (0, 0) \)
2. \( (0, \frac{7}{3}) \)
3. \( (\frac{7}{2}, 0) \)

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 4x + 45y \) at each corner point:
1. At \( (0, 0) \):
\[ Z = 4(0) + 45(0) = 0 \]
2. At \( (0, \frac{7}{3}) \):
\[ Z = 4(0) + 45\left(\frac{7}{3}\right) = 0 + 105 = 105 \]
3. At \( (\frac{7}{2}, 0) \):
\[ Z = 4\left(\frac{7}{2}\right) + 45(0) = 14 + 0 = 14 \]

### Step 8: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (0, \frac{7}{3}) \) with \( Z = 105 \).

### Final Answer
The maximum value of \( Z \) is \( \boxed{105} \) at the point \( (0, \frac{7}{3}) \).

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