MHT CET · Maths · Application of Derivatives
A circular sector of perimeter 60 meter with maximum area is to be constructed. The radius of the circular are in meter must be
- A \(5 \mathrm{~m}\)
- B \(15 \mathrm{~m}\)
- C \(10 \mathrm{~m}\)
- D \(20 \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(15 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Perimeter \(=2 r+r \theta=60\)
\(\Rightarrow \theta=\frac{60-2 r}{r}\)
Area \(=\mathrm{A}(\mathrm{r})=\frac{\pi \mathrm{r}^2 \theta}{360}=\frac{\pi \mathrm{r}^2\left(\frac{60-2 \mathrm{r}}{\mathrm{r}}\right)}{360}=\frac{\pi}{360}\left(60 r-2 \mathrm{r}^2\right)\)
\(A^{\prime}(r)=\frac{\pi}{360}(60-4 r)\)
For area to be maximum \(A^{\prime}(r)=0 \Rightarrow r=15\)
\(\Rightarrow \theta=\frac{60-2 r}{r}\)
Area \(=\mathrm{A}(\mathrm{r})=\frac{\pi \mathrm{r}^2 \theta}{360}=\frac{\pi \mathrm{r}^2\left(\frac{60-2 \mathrm{r}}{\mathrm{r}}\right)}{360}=\frac{\pi}{360}\left(60 r-2 \mathrm{r}^2\right)\)
\(A^{\prime}(r)=\frac{\pi}{360}(60-4 r)\)
For area to be maximum \(A^{\prime}(r)=0 \Rightarrow r=15\)
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