From Fig 6 16,
we have
RP2 = AR2 + AP2
and
RQ2 = RB2 + BQ2
Therefore
RP2 + RQ2 = AR2 + AP2 + RB2 + BQ2
= x2 + (16)2 + (20 – x)2 + (22)2
= 2x2 – 40x + 1140
Let
S ≡ S(x) = RP2 + RQ2 = 2x2 – 40x + 1140 Therefore
S′(x) = 4x – 40 Now S′(x) = 0 gives x = 10