The point for which the maximum value of z=x+y subject to the constraints `2x+5yle100,(x)/(25)+(y)/(50)le1,xge0,yge0` is obtained at

Options

A. (10,20)
B. (25,0)
C. (0,20)
D. `((75)/(4),(25)/(2))`

Answer

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

### Step 1: Identify the Constraints
The constraints given are:
1. \( 2x + 5y \leq 100 \)
2. \( \frac{x}{25} + \frac{y}{50} \leq 1 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Constraints to Equations
We will convert the inequalities to equations to find the boundary lines:
1. From \( 2x + 5y = 100 \)
2. From \( \frac{x}{25} + \frac{y}{50} = 1 \) which simplifies to \( 2x + y = 100 \)

### Step 3: Find Intercepts
For the first equation \( 2x + 5y = 100 \):
- Set \( x = 0 \): \( 5y = 100 \) → \( y = 20 \) (intercept at (0, 20))
- Set \( y = 0 \): \( 2x = 100 \) → \( x = 50 \) (intercept at (50, 0))

For the second equation \( 2x + y = 100 \):
- Set \( x = 0 \): \( y = 100 \) (intercept at (0, 100))
- Set \( y = 0 \): \( 2x = 100 \) → \( x = 50 \) (intercept at (50, 0))

### Step 4: Graph the Constraints
Plot the lines on a graph:
- The line \( 2x + 5y = 100 \) intersects the axes at (0, 20) and (50, 0).
- The line \( 2x + y = 100 \) intersects the axes at (0, 100) and (50, 0).

### Step 5: Identify the Feasible Region
The feasible region is bounded by the axes and the lines, considering the inequalities. The region is below both lines and in the first quadrant.

### Step 6: Find Corner Points
The corner points of the feasible region can be found by:
1. Intersection of \( 2x + 5y = 100 \) and \( 2x + y = 100 \).
- Solve the equations:
\[ 2x + 5y = 100 \quad (1) \]
\[ 2x + y = 100 \quad (2) \]
- Subtract (2) from (1):
\[ 4y = 0 \quad \Rightarrow \quad y = 0 \]
- Substitute \( y = 0 \) into (2):
\[ 2x = 100 \quad \Rightarrow \quad x = 50 \]
- Thus, one corner point is (50, 0).

2. Points from intercepts:
- (0, 20)
- (25, 0)
- (75/4, 25/2)

### Step 7: Evaluate \( z = x + y \) at Each Corner Point
1. At (0, 0): \( z = 0 + 0 = 0 \)
2. At (0, 20): \( z = 0 + 20 = 20 \)
3. At (25, 0): \( z = 25 + 0 = 25 \)
4. At \( (75/4, 25/2) \):
\[ z = \frac{75}{4} + \frac{25}{2} = \frac{75}{4} + \frac{50}{4} = \frac{125}{4} = 31.25 \]

### Step 8: Determine Maximum Value
The maximum value of \( z \) occurs at the point \( \left( \frac{75}{4}, \frac{25}{2} \right) \) with \( z = 31.25 \).

### Conclusion
The point for which the maximum value of \( z = x + y \) is obtained is \( \left( \frac{75}{4}, \frac{25}{2} \right) \).

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