Poles AP$=16$ m and BQ$=22$ m stand at A and B with AB$=20$ m. A point R on AB at distance $x$ from A minimises $RP^2+RQ^2$. The value of $x$ is

1. $RP^2+RQ^2=x^2+16^2+(20-x)^2+22^2$. 2. $S(x)=2x^2-40x+1140$. 3. $S'(x)=4x-40=0 \Rightarrow x=10$. 4. $S''(x)=4>0$, so $x=10$ gives the minimum. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.21_