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 $a$.

• A. $breadth=\frac{2a}{\left(3\right)}$ $depth=4a\frac{\sqrt{2}}{3}$
• B. $breadth=\frac{a}{\sqrt{\left(3\right)}}$ $depth=a\sqrt{\left(\frac{11}{3}\right)}.$
• C. $breadth=\frac{2a}{\sqrt{\left(3\right)}}$ $depth=2a\sqrt{\left(\frac{2}{3}\right)}.$
• D. $breadth=\frac{a}{\sqrt{\left(2\right)}}$ $depth=2a\sqrt{\left(\frac{7}{2}\right)}.$

Answer 1

The correct option is C $breadth=\frac{2a}{\sqrt{\left(3\right)}}$ $depth=2a\sqrt{\left(\frac{2}{3}\right)}.$

Let the breadth of the beam be $x$ and depth be $y$
Then, $x^2+y^2=4a^2$
Strength $=S=kx{y}^2$ given
or $S=kx(4a^2-x^2)=k(4a^{2}x-x^3).$
For maxima or minima,
$\frac{dS}{dx}=0$
$\Rightarrow k(4a^2-3x^2)=0$
$\Rightarrow x=\frac{2a}{\sqrt{3}}$
Now, $\frac{d^{2}S}{d{x}^2}=-6kx<0$ at $x=\frac{2a}{\sqrt{3}}$
Hence, $S$ is maximum when $x=\frac{2a}{\sqrt{3}}$
and therefore $y=2a\sqrt{\left(\frac{2}{3}\right)}.$