Question

Let the points \( \left( \frac{11}{2}, \alpha \right) \) lie on or inside the triangle with sides \( x + y = 11 \), \( x + 2y = 16 \), and \( 2x + 3y = 29 \). Then the product of the smallest and the largest values of \( \alpha \) is equal to:

• A. \( 55 \)
• B. \( 33 \)
• C. \( 22 \)
• D. \( 44 \)

Hint:

To solve for the values of \( \alpha \), find the points of intersection of the lines forming the triangle and check the conditions for the points to be inside or on the boundary of the triangle.

Answer 1

The Correct Option is B

We are given the points \( \left( \frac{11}{2}, \alpha \right) \) that lie inside or on the boundary of the triangle formed by the lines \( x + y = 11 \), \( x + 2y = 16 \), and \( 2x + 3y = 29 \).

Step 1: Find the equation of the triangle

We first solve the system of equations for the lines forming the triangle.

• The line \( x + y = 11 \) is the first boundary.
• The second line \( x + 2y = 16 \) intersects the first line at a point we need to find.
• The third line \( 2x + 3y = 29 \) intersects the first two lines at another set of points.

Step 2: Solve for the points of intersection

We solve these systems of linear equations to find the boundaries of the triangle and determine the limits for \( \alpha \), the y-coordinate of the point \( \left( \frac{11}{2}, \alpha \right) \).

The values of \( \alpha \) that satisfy the condition for the points to lie inside or on the triangle will give the smallest and largest values of \( \alpha \).

Step 3: Find the product of the smallest and largest values of \( \alpha \)

Once the smallest and largest values of \( \alpha \) are identified, we compute their product.

After solving, we find that the product of the smallest and largest values of \( \alpha \) is \( 33 \).

Final Answer: \( 33 \).

Answer 2

Step 1: Given lines are:
L₁: x + y = 11
L₂: x + 2y = 16
L₃: 2x + 3y = 29

Step 2: Find the points of intersection (vertices of the triangle).

(a) Intersection of L₁ and L₂:
Subtract → (x + y) − (x + 2y) = 11 − 16 ⇒ −y = −5 ⇒ y = 5
Substitute y = 5 in L₁: x + 5 = 11 ⇒ x = 6
Hence, A(6, 5)

(b) Intersection of L₁ and L₃:
From L₁: y = 11 − x
Substitute in L₃: 2x + 3(11 − x) = 29 ⇒ 2x + 33 − 3x = 29 ⇒ −x = −4 ⇒ x = 4
Then y = 11 − 4 = 7
Hence, B(4, 7)

(c) Intersection of L₂ and L₃:
From L₂: x = 16 − 2y
Substitute in L₃: 2(16 − 2y) + 3y = 29 ⇒ 32 − 4y + 3y = 29 ⇒ −y = −3 ⇒ y = 3
Then x = 16 − 2(3) = 10
Hence, C(10, 3)

Therefore, the vertices of the triangle are A(6, 5), B(4, 7), and C(10, 3).

Step 3: Consider the vertical line x = 11/2 = 5.5.
Find the corresponding y-values on each line.

L₁: y = 11 − x = 11 − 5.5 = 5.5
L₂: y = (16 − x)/2 = (16 − 5.5)/2 = 10.5/2 = 5.25
L₃: y = (29 − 2x)/3 = (29 − 11)/3 = 18/3 = 6

Step 4: The line x = 5.5 cuts the triangle between points on L₁ and L₃.
So, α varies from the smallest value 5.5 to the largest value 6.

Step 5: Product of smallest and largest α values:
= 5.5 × 6 = (11/2) × 6 = 33

Final Answer: 33