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

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

Answers (1)

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

 

Posted by

infoexpert24

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