Question

The strength of a beam varies as the product of its breadth $b$ and square of its depth $d.$ A beam cut out of a circular log of radius $r$ would be strong when

• A. $b=d=\frac{r}{2}$
• B. $b^2=\frac{r}{2\sqrt{2}}=d$
• C. $d=\sqrt{2}b=\sqrt{\frac{2}{3}}\cdot 2r$
• D. $d=\sqrt{3}b=\sqrt{\frac{3}{2}}\cdot 2r$

Answer 1

The correct option is C $d=\sqrt{2}b=\sqrt{\frac{2}{3}}\cdot 2r$

Strength of the beam $S=kb{d}^2$
$\frac{d^2}{4}+\frac{b^2}{4}=r^2$
$\Rightarrow d^2=4r^2-b^2$
$\therefore S=kb(4r^2-b^2)=4kb{r}^2-k{b}^3$
$\frac{dS}{db}=4k{r}^2-3k{b}^2,\frac{d^{2}S}{d{b}^2}=-6kb<0$
$\frac{dS}{db}=0$
$\Rightarrow b^2=\frac{4}{3}{r}^2\Rightarrow b=\frac{2}{\sqrt{3}}r$
$d^2=4r^2-\frac{4r^2}{3}=\frac{8r^2}{3}$
$\Rightarrow d=\sqrt{\frac{2}{3}}\cdot 2r$