MHT CET · Maths · Application of Derivatives
20 meters of wire is available to fence of a flowerbed in the form of a circular sector. If the flowerbed is to have maximum surface area, then the radius of the circle is
- A \(8 \mathrm{~m}\)
- B \(4 \mathrm{~m}\)
- C \(2 \mathrm{~m}\)
- D \(5 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(5 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\theta^c=\frac{\ell}{r}=\frac{20-2 r}{r}=\frac{20-2 r}{r} \times \frac{180^{\circ}}{\pi}\)
Now area of sector \(=\frac{\pi r^2 \theta}{360^{\circ}}\)
\(A(r)=\frac{\pi r^2 \times \frac{20-2 r}{r} \times \frac{180^{\circ}}{\pi}}{360^{\circ}}=10 r-r^2\)
For maximum area \(A^{\prime}(r)=0\)
\(\begin{aligned} & \Rightarrow 10-2 r=0 \\ & \Rightarrow r=5 \mathrm{~m}\end{aligned}\)
Now area of sector \(=\frac{\pi r^2 \theta}{360^{\circ}}\)
\(A(r)=\frac{\pi r^2 \times \frac{20-2 r}{r} \times \frac{180^{\circ}}{\pi}}{360^{\circ}}=10 r-r^2\)
For maximum area \(A^{\prime}(r)=0\)
\(\begin{aligned} & \Rightarrow 10-2 r=0 \\ & \Rightarrow r=5 \mathrm{~m}\end{aligned}\)
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