The strength of a beam varies as the product of its breadth and square of its depth.

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:

Let the breadth, height and strength of the beam be $b, h$ and $S$, respectively.

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

$\Rightarrow 4 a^{2}-b^{2}=h^{2}$             ....(1)

Here,

Strength of beam, $S=K b h^{2}$

$\Rightarrow S=k b\left(4 R^{2}-b^{2}\right)$           [Where $K$ is some constant]

$\Rightarrow S=k\left(b 4 a^{2}-b^{3}\right)$        $[$ From eq. (1) $]$

$\Rightarrow \frac{d S}{d b}=k\left(4 a^{2}-3 b^{2}\right)$

For maximum or minimum values of $S$, we must have

$\frac{d S}{d b}=0$

$\Rightarrow k\left(4 a^{2}-3 b^{2}\right)=0$

$\Rightarrow 4 a^{2}-3 b^{2}=0$

$\Rightarrow 4 a^{2}=3 b^{2}$

$\Rightarrow b=\frac{2 a}{\sqrt{3}}$

Substituting the value of $b$ in eq. (1), we get

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

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

$\Rightarrow h=\frac{2 \sqrt{2}}{\sqrt{3}} a$

Now,

$\frac{d^{2} S}{d b^{2}}=-6 K b$

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

$\Rightarrow \frac{d^{2} S}{d b^{2}}=\frac{-12 K a}{\sqrt{3}}<0$

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

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