If the sum of the heights of transmitting and receiving antennas in line of sight of communication is fixed at h, show that the range is maximum when the two antennas have a height `h//2` each.

Options

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

Explanation

To show that the range is maximum when the heights of the transmitting and receiving antennas are both \( \frac{h}{2} \), we can follow these steps:

### Step 1: Define the Variables
Let:
- \( H_t \) = height of the transmitting antenna
- \( H_r \) = height of the receiving antenna
- \( h \) = fixed sum of the heights, so \( H_t + H_r = h \)

### Step 2: Write the Expression for Range
The range \( R \) of line-of-sight communication is given by the formula:
\[ R = \sqrt{2 H_t} + \sqrt{2 H_r} \]

### Step 3: Substitute the Condition
Since \( H_r = h - H_t \), we can substitute this into the range equation:
\[ R = \sqrt{2 H_t} + \sqrt{2 (h - H_t)} \]

### Step 4: Differentiate the Range with Respect to \( H_t \)
To find the maximum range, we need to differentiate \( R \) with respect to \( H_t \) and set the derivative equal to zero:
\[ \frac{dR}{dH_t} = \frac{1}{\sqrt{2 H_t}} - \frac{1}{\sqrt{2 (h - H_t)}} = 0 \]

### Step 5: Solve the Equation
Setting the derivative equal to zero gives:
\[ \frac{1}{\sqrt{2 H_t}} = \frac{1}{\sqrt{2 (h - H_t)}} \]
Squaring both sides leads to:
\[ 2 (h - H_t) = 2 H_t \]
This simplifies to:
\[ h - H_t = H_t \]
Thus:
\[ h = 2 H_t \quad \Rightarrow \quad H_t = \frac{h}{2} \]

### Step 6: Find the Height of the Receiving Antenna
Using the condition \( H_r = h - H_t \):
\[ H_r = h - \frac{h}{2} = \frac{h}{2} \]

### Conclusion
Thus, both antennas should have a height of \( \frac{h}{2} \) for the range to be maximum.

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