A tree trunk  6 metre  long is in the shape of a freestum of a cone, the radii of its ends  being 4 and 3 metres. It is required to cut from it a beam of uniform  square cross section . Then the height of the bean of maximum volume is.

Let 2x metre be the length of the diagonal of the square  cross  section of the bean and let h metre be its height .

$$\begin{gathered} oA=oB=4,o\text{'}c=o\text{'}d=3,LN=2x \\ letvbethevolumeofthebean \\ v=Areaofbasexheight=\frac{1}{2}{\left(diagonal\right)}^{2}xheight \\ =\frac{1}{2}{\left(CN\right)}^{2}xpL=\frac{1}{2}{\left(2x\right)}^2.h =2x^{2}h \end{gathered}$$

$CS\perp AB$

$$\begin{gathered} triangleBCSandBPLaresimilar \\ \frac{CS}{PL}=\frac{BS}{BL} \\ \frac{6}{h}=\frac{4-3}{4-x} \\ h=6(h-x) \end{gathered}$$