The sum of the heights of transmitting and receiving attennas in line of sight of communication is fixed at h, find the height of two antennas when r ange is maximum.

Options

A. `h//2`
B. 2h
C. h
D. 4h

Explanation

To solve the problem of finding the heights of the transmitting and receiving antennas when the range is maximized, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
Let \( H_T \) be the height of the transmitting antenna and \( H_R \) be the height of the receiving antenna. According to the problem, we have:
\[ H_T + H_R = h \]

2. **Express One Variable in Terms of the Other**:
From the equation above, we can express \( H_R \) in terms of \( H_T \):
\[ H_R = h - H_T \]

3. **Write the Formula for Range**:
The range \( R \) of line-of-sight communication between the two antennas is given by:
\[ R = 2 \sqrt{H_T} + 2 \sqrt{H_R} \]
Substituting \( H_R \) from step 2:
\[ R = 2 \sqrt{H_T} + 2 \sqrt{h - H_T} \]

4. **Maximize the Range**:
To find the maximum range, we need to differentiate \( R \) with respect to \( H_T \) and set the derivative to zero:
\[ \frac{dR}{dH_T} = 2 \cdot \frac{1}{2\sqrt{H_T}} - 2 \cdot \frac{1}{2\sqrt{h - H_T}} = 0 \]
Simplifying this gives:
\[ \frac{1}{\sqrt{H_T}} = \frac{1}{\sqrt{h - H_T}} \]

5. **Solve for \( H_T \)**:
Squaring both sides results in:
\[ h - H_T = H_T \]
Rearranging gives:
\[ h = 2H_T \]
Therefore:
\[ H_T = \frac{h}{2} \]

6. **Find \( H_R \)**:
Using the relationship \( H_R = h - H_T \):
\[ H_R = h - \frac{h}{2} = \frac{h}{2} \]

### Conclusion:
Thus, the heights of both the transmitting and receiving antennas for maximum range are:
\[ H_T = \frac{h}{2}, \quad H_R = \frac{h}{2} \]