Maximize `Z = 400x + 500y` subject to constraints `x + 2y le 80 ,2x + y le 90, x ge 0 , y ge 0`

Answer

To solve the problem of maximizing \( Z = 400x + 500y \) subject to the given constraints, we will follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + 2y \leq 80 \)
2. \( 2x + y \leq 90 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines of the constraints, we convert the inequalities into equations:
1. \( x + 2y = 80 \)
2. \( 2x + y = 90 \)

### Step 3: Find Intercepts of the Lines
For the first line \( x + 2y = 80 \):
- When \( x = 0 \): \( 2y = 80 \) → \( y = 40 \) (Point: \( (0, 40) \))
- When \( y = 0 \): \( x = 80 \) (Point: \( (80, 0) \))

For the second line \( 2x + y = 90 \):
- When \( x = 0 \): \( y = 90 \) (Point: \( (0, 90) \))
- When \( y = 0 \): \( 2x = 90 \) → \( x = 45 \) (Point: \( (45, 0) \))

### Step 4: Find the Intersection of the Lines
To find the intersection of the two lines, we solve the equations:
1. \( x + 2y = 80 \)
2. \( 2x + y = 90 \)

From the first equation, express \( x \):
\[ x = 80 - 2y \]

Substituting into the second equation:
\[ 2(80 - 2y) + y = 90 \]
\[ 160 - 4y + y = 90 \]
\[ 160 - 3y = 90 \]
\[ 3y = 70 \]
\[ y = \frac{70}{3} \]

Now substitute \( y \) back to find \( x \):
\[ x = 80 - 2\left(\frac{70}{3}\right) \]
\[ x = 80 - \frac{140}{3} \]
\[ x = \frac{240 - 140}{3} = \frac{100}{3} \]

Thus, the intersection point is \( \left(\frac{100}{3}, \frac{70}{3}\right) \).

### Step 5: Identify the Feasible Region
The feasible region is determined by the constraints. We plot the lines and shade the area that satisfies all inequalities:
- The area must be below both lines and in the first quadrant (since \( x \geq 0 \) and \( y \geq 0 \)).

### Step 6: Evaluate the Objective Function at the Vertices
The vertices of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 40) \)
3. \( (45, 0) \)
4. \( \left(\frac{100}{3}, \frac{70}{3}\right) \)

Now we calculate \( Z \) at each vertex:
1. At \( (0, 0) \):
\[ Z = 400(0) + 500(0) = 0 \]

2. At \( (0, 40) \):
\[ Z = 400(0) + 500(40) = 20000 \]

3. At \( (45, 0) \):
\[ Z = 400(45) + 500(0) = 18000 \]

4. At \( \left(\frac{100}{3}, \frac{70}{3}\right) \):
\[ Z = 400\left(\frac{100}{3}\right) + 500\left(\frac{70}{3}\right) \]
\[ = \frac{40000}{3} + \frac{35000}{3} = \frac{75000}{3} \approx 25000 \]

### Step 7: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( \left(\frac{100}{3}, \frac{70}{3}\right) \) and is:
\[ Z_{\text{max}} = 25000 \]

### Final Answer
Thus, the maximum value of \( Z \) is \( 25000 \) at the point \( \left(\frac{100}{3}, \frac{70}{3}\right) \).

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