#### Question

The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius a ?

#### Solution

\[\text { Let the breadth, height and strength of the beam be b, h and S, respectively }. \]

\[ a^2 = \frac{h^2 + b^2}{4}\]

\[ \Rightarrow 4 a^2 - b^2 = h^2 .............. \left( 1 \right)\]

\[\text { Here }, \]

\[\text { Strength of beam,} S = Kb h^2 ............\left[ \text { Where K is some constant } \right]\]

\[ \Rightarrow S = kb\left( 4 R^2 - b^2 \right) .................\left[ \text { From eq. }\left( 1 \right) \right]\]

\[ \Rightarrow S = k\left( b4 a^2 - b^3 \right)\]

\[ \Rightarrow \frac{dS}{db} = k\left( 4 a^2 - 3 b^2 \right)\]

\[\text { For maximum or minimum values of S, we must have }\]

\[\frac{dS}{db} = 0\]

\[ \Rightarrow k\left( 4 a^2 - 3 b^2 \right) = 0\]

\[ \Rightarrow 4 a^2 - 3 b^2 = 0\]

\[ \Rightarrow 4 a^2 = 3 b^2 \]

\[ \Rightarrow b = \frac{2a}{\sqrt{3}}\]

\[\text { Substituting the value of b in eq }. \left( 1 \right),\text { we get }\]

\[ \Rightarrow 4 a^2 - \left( \frac{2a}{\sqrt{3}} \right)^2 = h^2 \]

\[ \Rightarrow \frac{12 a^2 - 4 a^2}{3} = h^2 \]

\[ \Rightarrow h = \frac{2\sqrt{2}}{\sqrt{3}}a\]

\[\text { Now, }\]

\[\frac{d^2 S}{d b^2} = - 6Kb\]

\[ \Rightarrow \frac{d^2 S}{d b^2} = - 6K\frac{2a}{\sqrt{3}}\]

\[ \Rightarrow \frac{d^2 S}{d b^2} = \frac{- 12Ka}{\sqrt{3}} < 0\]

\[\text { So, the strength of beam is maximum when b =} \frac{2a}{\sqrt{3}} \text { and h } = \frac{2\sqrt{2}}{\sqrt{3}}a . \]