#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 42.

$b= \frac{2a}{\sqrt{3}}, h=\frac{2\sqrt{2}}{\sqrt{3}}a$

Hint: For maximum or minimum value of S, we must have $\frac{dS}{dD}=0$

Given:

Let the breadth, height, strength of the beam be b,h,S

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

$4a^2-b^2=h^2$                      .......(1)

Solution:

Stength beam, $S=kbh^2$

$S=kb(4a^2-b^2)$                       ......from (1)

$S=k(b4a^2-b^3)$

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

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

\begin{aligned} &k\left(4 a^{2}-3 b^{2}\right)=0 \\ &4 a^{2}-3 b^{2}=0 \\ &4 a^{2}=3 b^{2} \\ &b=\frac{2 a}{\sqrt{3}} \end{aligned}

Sub b value in (1)

\begin{aligned} &4 a^{2}-\left(\frac{2 a}{\sqrt{3}}\right)^{2}=h^{2} \\ &\frac{12 a^{2}-4 a^{2}}{3}=h^{2} \\ &h=\frac{2 \sqrt{2}}{\sqrt{3}} a \end{aligned}

\begin{aligned} &\frac{d^{2} S}{d b^{2}}=-6 k b \\ &=-6 k \frac{2 a}{\sqrt{3}} \\ &=\frac{-12 k a}{\sqrt{3}}<0 \end{aligned}

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