The point of which of the maximum value of ` z= x+y` subject to contraints ` x+2yle 70,2x+yle 90,x ge 0,yge0` is obtained at

Options

A. (30,25)
B. (20,35)
C. (35,20)
D. `(110/3, 50/3)`

Answer

To find the maximum value of \( z = x + y \) subject to the constraints \( x + 2y \leq 70 \), \( 2x + y \leq 90 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

### Step 1: Identify the Constraints
The constraints are:
1. \( x + 2y \leq 70 \)
2. \( 2x + y \leq 90 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x + 2y = 70 \)
2. \( 2x + y = 90 \)

### Step 3: Find Intercepts of Each Line
**For the first line \( x + 2y = 70 \):**
- Set \( y = 0 \):
\[ x + 2(0) = 70 \implies x = 70 \quad \text{(Point: (70, 0))} \]
- Set \( x = 0 \):
\[ 0 + 2y = 70 \implies y = 35 \quad \text{(Point: (0, 35))} \]

**For the second line \( 2x + y = 90 \):**
- Set \( y = 0 \):
\[ 2x + 0 = 90 \implies x = 45 \quad \text{(Point: (45, 0))} \]
- Set \( x = 0 \):
\[ 2(0) + y = 90 \implies y = 90 \quad \text{(Point: (0, 90))} \]

### Step 4: Find the Intersection of the Two Lines
To find the intersection, we solve the equations:
1. From \( x + 2y = 70 \), we can express \( x \) in terms of \( y \):
\[ x = 70 - 2y \]
2. Substitute \( x \) in the second equation \( 2x + y = 90 \):
\[ 2(70 - 2y) + y = 90 \]
\[ 140 - 4y + y = 90 \]
\[ 140 - 3y = 90 \implies 3y = 50 \implies y = \frac{50}{3} \]
3. Substitute \( y \) back to find \( x \):
\[ x = 70 - 2\left(\frac{50}{3}\right) = 70 - \frac{100}{3} = \frac{210 - 100}{3} = \frac{110}{3} \]

### Step 5: Identify the Feasible Region
The feasible region is bounded by the points:
- \( (70, 0) \)
- \( (0, 35) \)
- \( (45, 0) \)
- \( (0, 90) \)
- \( \left(\frac{110}{3}, \frac{50}{3}\right) \)

### Step 6: Evaluate \( z = x + y \) at Each Vertex
1. At \( (0, 35) \):
\[ z = 0 + 35 = 35 \]
2. At \( (45, 0) \):
\[ z = 45 + 0 = 45 \]
3. At \( \left(\frac{110}{3}, \frac{50}{3}\right) \):
\[ z = \frac{110}{3} + \frac{50}{3} = \frac{160}{3} \approx 53.33 \]

### Step 7: Determine the Maximum Value
The maximum value of \( z \) occurs at the point \( \left(\frac{110}{3}, \frac{50}{3}\right) \) with \( z \approx 53.33 \).

### Conclusion
The point at which the maximum value of \( z = x + y \) is obtained is:
\[ \left(\frac{110}{3}, \frac{50}{3}\right) \]