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

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

Given: m=\frac{WL}{2}x-\frac{W}{2}x^2

Solution: m=\frac{WL}{2}x-\frac{W}{2}x^2

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

For maximum or minimum ,






So m is maximum at x=\frac{L}{2}

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