Consider point A(6, 30), point B(24, 6) and line AB: 4x+3y = 114.
Point `P(0, lambda)` is a point on y-axis such that `0 lt lambda lt 38 " and point " Q(0, lambda)` is a point on y-axis such that `lambda gt 38`.
For all positions of pont Q, and AQB is maximum when point Q is

Options

A. (0, 54)
B. (0, 58)
C. (0, 60)
D. (0, 1)

Explanation

To solve the problem step by step, we will analyze the given points and the line, and then find the position of point Q on the y-axis such that the angle AQB is maximized.

### Step 1: Identify the points and the line
We have:
- Point A: \( A(6, 30) \)
- Point B: \( B(24, 6) \)
- Line AB: \( 4x + 3y = 114 \)

### Step 2: Determine the coordinates of point P and Q
Point P is defined as \( P(0, \lambda) \) where \( 0 < \lambda < 38 \) and point Q is defined as \( Q(0, \lambda) \) where \( \lambda > 38 \).

### Step 3: Find the angle AQB
To find the angle \( \angle AQB \), we can use the tangent of the angle formula. The tangent of the angle \( \theta \) can be expressed in terms of the coordinates of points A, B, and Q.

Using the coordinates:
- \( A(6, 30) \)
- \( B(24, 6) \)
- \( Q(0, \lambda) \)

The tangent of the angle \( \angle AQB \) can be calculated as:
\[ \tan(\theta) = \frac{(y_Q - y_A)(x_B - x_A)}{(x_B - x_A)(y_Q - y_B) - (y_B - y_A)(x_Q - x_A)} \]
Substituting the coordinates, we can simplify this expression.

### Step 4: Differentiate to find maximum
To find the maximum angle \( AQB \), we need to differentiate \( \tan(\theta) \) with respect to \( \lambda \) and set the derivative equal to zero.

From the video transcript, we have:
\[ \tan(\theta) = \frac{(38 - \lambda)(18)}{(\lambda - 18)^2} \]
Taking the derivative:
\[ \frac{d(\tan(\theta))}{d\lambda} = \frac{(18 - \lambda)(\lambda - 58)}{(\lambda - 18)^4} \]
Setting this equal to zero gives us critical points.

### Step 5: Solve for critical points
Setting \( (18 - \lambda)(\lambda - 58) = 0 \) gives us:
- \( \lambda = 18 \)
- \( \lambda = 58 \)

### Step 6: Determine the maximum
To determine which of these points gives a maximum angle, we can analyze the sign of the derivative around these points. The derivative changes from positive to negative at \( \lambda = 58 \), indicating that this is a maximum.

### Step 7: Conclusion
Thus, the point Q that maximizes the angle \( AQB \) is:
\[ Q(0, 58) \]