Consider `x/2 + y/4 gt= 1 and x/3 + y/2 lt=1 ,x,y gt=0` number of possible solutions are :

Options

A. zero
B. unique
C. infinite
D. none of these

Answer

To solve the problem, we need to analyze the inequalities given and find the feasible region defined by them. Let's break down the solution step by step:

### Step 1: Write the inequalities
We have two inequalities:
1. \(\frac{x}{2} + \frac{y}{4} \geq 1\)
2. \(\frac{x}{3} + \frac{y}{2} \leq 1\)

### Step 2: Convert inequalities to standard form
To make it easier to work with, we can multiply through by the least common multiple (LCM) of the denominators to eliminate the fractions.

For the first inequality:
\[ \frac{x}{2} + \frac{y}{4} \geq 1 \implies 2x + y \geq 4 \]

For the second inequality:
\[ \frac{x}{3} + \frac{y}{2} \leq 1 \implies 2x + 3y \leq 6 \]

### Step 3: Identify the boundaries
Now we have the following two boundary lines:
1. \(2x + y = 4\)
2. \(2x + 3y = 6\)

### Step 4: Find intercepts for both lines
**For the first line \(2x + y = 4\):**
- If \(x = 0\), then \(y = 4\) (y-intercept).
- If \(y = 0\), then \(2x = 4 \implies x = 2\) (x-intercept).

**For the second line \(2x + 3y = 6\):**
- If \(x = 0\), then \(3y = 6 \implies y = 2\) (y-intercept).
- If \(y = 0\), then \(2x = 6 \implies x = 3\) (x-intercept).

### Step 5: Plot the lines on a graph
Plot the points:
- For \(2x + y = 4\), plot (0, 4) and (2, 0).
- For \(2x + 3y = 6\), plot (0, 2) and (3, 0).

### Step 6: Determine the feasible region
Now we need to determine which side of each line satisfies the inequality:
- For \(2x + y \geq 4\), test the point (0,0): \(2(0) + 0 \geq 4\) is false, so the feasible region is above the line.
- For \(2x + 3y \leq 6\), test the point (0,0): \(2(0) + 3(0) \leq 6\) is true, so the feasible region is below this line.

### Step 7: Identify the intersection point
To find the intersection of the two lines, set the equations equal to each other:
\[ 2x + y = 4 \quad (1) \]
\[ 2x + 3y = 6 \quad (2) \]

From equation (1), we can express \(y\) in terms of \(x\):
\[ y = 4 - 2x \]

Substituting \(y\) in equation (2):
\[ 2x + 3(4 - 2x) = 6 \]
\[ 2x + 12 - 6x = 6 \]
\[ -4x + 12 = 6 \implies -4x = -6 \implies x = \frac{3}{2} \]

Now substitute \(x = \frac{3}{2}\) back into equation (1):
\[ 2\left(\frac{3}{2}\right) + y = 4 \implies 3 + y = 4 \implies y = 1 \]

So the intersection point is \(\left(\frac{3}{2}, 1\right)\).

### Step 8: Describe the feasible region
The feasible region is bounded by the lines and the axes, and it includes all points where both inequalities are satisfied. Since both \(x\) and \(y\) must be greater than or equal to 0, we consider only the first quadrant.

### Step 9: Conclusion on the number of solutions
The feasible region is a polygon (specifically a triangle) with vertices at (0, 4), (2, 0), (3, 0), and (0, 2). Since this region contains infinitely many points, the number of possible solutions is infinite.

### Final Answer
The number of possible solutions is **infinite**.