If the sum of the heights of transmitting and

$\mathrm{h}_{\mathrm{T}}=\mathrm{h}_{\mathrm{R}}=160 \ldots$ (i)

$\mathrm{d}=\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}+\sqrt{2 \mathrm{Rh}_{\mathrm{R}}}$

$d=\sqrt{2 R}\left[\sqrt{h_{T}}+\sqrt{h_{R}}\right]$

$d=\sqrt{2 R}[\sqrt{x}+\sqrt{160-x}]$

$\frac{d(d)}{d x}=0$

$\frac{1}{2 \sqrt{x}}+\frac{1(-1)}{2 \sqrt{160-x}}=0$

$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{160-x}}$

$x=80 m$

$\mathrm{d}_{\max }=\sqrt{2 \times 6400}\left[\sqrt{\frac{80}{1000}}+\sqrt{\frac{20}{1000}}\right]$'

$=\frac{80 \sqrt{2} \times 2 \sqrt{80}}{10 \sqrt{10}}$

$=8 \times 2 \times \sqrt{2} \times 2 \sqrt{2}=64 \mathrm{~km}$