PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# 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? - PUC Karnataka Science Class 12 - Mathematics

#### 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

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

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

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

$\text { Here },$

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

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

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

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

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

$\frac{dS}{db} = 0$

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

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

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

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

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

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

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

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

$\text { Now, }$

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

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

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

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

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Solution 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? Concept: Graph of Maxima and Minima.
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