A 3' cube sits on level ground against a vertical wall. A 12' ladder on the same ground leans against the wall such that it touches the top edge of the box.

How far from the wall must the foot of the ladder be, if it is to reach maximum height whilst meeting the foregoing conditions?

(In reply to Solution by Brian Smith)

Let the overall triangle have sides U and V (and hypontenuse 12) so that U=12cosT, V=12sinT. Then by similar triangles, (U-3)/3=3/(V-3) or UV-3U-3V=0 or 4sinT=1+tanT. Squaring both sides and multiplying by (cosT)^2 gives 16(sinTcosT)^2=1+2sinTcosT or 4(sin2T)^2=1+sin2T. Solving the quadratic gives sin2T=(1+sqrt17)/8 so that 2T=39.82077293 degrees. Thus U=12cosT=11.2827169 and V=12sinT=4.086599978.
Edited on December 9, 2003, 4:30 pm

Posted by Richard on 2003-12-09 16:26:15