Question

The area of the region \(\{ {(x, y): 0 ≤ x ≤ \frac{9}{4}, 0 ≤ y ≤ 1, x ≥ 3y, x + y ≥ 2}\}\)

• A. \(\frac{11}{32}\)
• B. \(\frac{35}{96}\)
• C. \(\frac{37}{96}\)
• D. \(\frac{13}{32}\)

Answer 1

The Correct Option is A

Step 1: Understanding the given region
The region is defined by the following conditions:
- \( 0 \leq x \leq \frac{9}{4} \)
- \( 0 \leq y \leq 1 \)
- \( x \geq 3y \)
- \( x + y \geq 2 \)
The constraints define a quadrilateral region bounded by the lines:
- \( x + y = 2 \)
- \( x = 3y \)
- \( y = 1 \)
- The vertical line \( x = \frac{9}{4} \)
Step 2: Identify the points of intersection
We first find the points where these lines intersect to determine the vertices of the region:
1. Intersection of \( x + y = 2 \) and \( y = 1 \):
- Substitute \( y = 1 \) into \( x + y = 2 \):
\[ x + 1 = 2 \implies x = 1 \] - So, the point of intersection is \( (1, 1) \).
2. Intersection of \( x + y = 2 \) and \( x = 3y \):
- Substitute \( x = 3y \) into \( x + y = 2 \):
\[ 3y + y = 2 \implies 4y = 2 \implies y = \frac{1}{2} \] - Now substitute \( y = \frac{1}{2} \) into \( x = 3y \):
\[ x = 3 \times \frac{1}{2} = \frac{3}{2} \] - So, the point of intersection is \( \left( \frac{3}{2}, \frac{1}{2} \right) \).
3. Intersection of \( x = 3y \) and \( y = 1 \):
- Substitute \( y = 1 \) into \( x = 3y \):
\[ x = 3 \times 1 = 3 \] - So, the point of intersection is \( (3, 1) \).
4. Intersection of the vertical line \( x = \frac{9}{4} \) and \( y = 1 \):
- This is straightforward: \( \left( \frac{9}{4}, 1 \right) \).
Step 3: Break the region into smaller parts and calculate the area
The region can be divided into smaller triangles, each having a base and height:
Triangle 1 (bounded by \( x = 3y \), \( x + y = 2 \), and \( y = 1 \)):
- Base of this triangle is \( x = 3 \) (from \( (1, 1) \) to \( (3, 1) \)).
- Height is \( \frac{1}{2} \) (from \( (1, 1) \) to \( \left( \frac{3}{2}, \frac{1}{2} \right) \)).
The area of this triangle is:
\[ A_1 = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 3 \times \frac{1}{2} = \frac{3}{4} \] Triangle 2 (bounded by \( x = 3y \), \( x = \frac{9}{4} \), and \( y = 1 \)):
- Base of this triangle is \( \frac{9}{4} - 3 = \frac{1}{4} \).
- Height is \( 1 \) (from \( (1, 1) \) to \( (3, 1) \)).
The area of this triangle is:
\[ A_2 = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times \frac{1}{4} \times 1 = \frac{1}{8} \] Triangle 3 (bounded by \( x + y = 2 \), \( x = 3y \), and \( y = 1 \)):
- Base of this triangle is \( 1 - \frac{3}{2} = \frac{1}{2} \).
- Height is \( 1 \) (from \( (1, 1) \) to \( \left( \frac{3}{2}, \frac{1}{2} \right) \)).
The area of this triangle is:
\[ A_3 = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4} \] Step 4: Calculate the total area
The total area of the required region is the sum of the areas of the three triangles: \[ A_{\text{total}} = A_1 + A_2 + A_3 = \frac{3}{4} + \frac{1}{8} + \frac{1}{4} = \frac{11}{32} \]