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Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 6 sub question (ii) maths textbook solution.

Answers (1)


Hint: For maximum or minimum value of we must have \frac{dm}{dx}=0

Given: m=\frac{Wx}{3}-\frac{Wx^3}{3L^2}

Solution: m=\frac{Wx}{3}-\frac{Wx^3}{3L^2}

\begin{aligned} &\frac{d m}{d x}=\frac{W}{3}-3 \times \frac{W x^{2}}{3 L^{2}} \\ &=\frac{W}{3}-\frac{W x^{2}}{L^{2}} \end{aligned}

For maximum or minimum,


\begin{aligned} &\frac{W}{3}-\frac{W x^{2}}{L^{2}}=0 \\ &\frac{W}{3}=\frac{W x^{2}}{L^{2}} \\ &x=\frac{L}{\sqrt{3}} \end{aligned}

\frac{d^{2} m}{d x^{2}}=\frac{-2 W x}{L^{2}}<0

So m is minimum at  x=\frac{L}{\sqrt{3}}

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