MHT CET · Maths · Application of Derivatives
20 meters wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible surface area, then the radius of the circle is
- A \(2 \mathrm{~m}\)
- B \(4 \mathrm{~m}\)
- C \(5 \mathrm{~m}\)
- D \(10 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(5 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Let \(\ell\) and \(\mathrm{r}\) be as shown in figure.
We have \(20=2 r+\ell\)
\(
\ell=20-2 \mathrm{r}
\)
Area of flower bed
\(
\begin{array}{l}
A=\frac{1}{2} \times \ell r=\frac{1}{2}(20-2 r) \cdot r \\
A=10 r-r^{2}
\end{array}
\)
\(
\begin{array}{l}
\frac{\mathrm{d} \mathrm{A}}{\mathrm{dr}}=10-2 \mathrm{r} \text { and } \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dr}^{2}}=-2 < 0 \\
\text { When } \frac{\mathrm{d} \mathrm{A}}{\mathrm{dr}}=0 \Rightarrow 10-2 \mathrm{r}=0 \Rightarrow \mathrm{r}=5
\end{array}
\)
Hence area of flower bed will be maximum when \(r=5\).
We have \(20=2 r+\ell\)
\(
\ell=20-2 \mathrm{r}
\)
Area of flower bed
\(
\begin{array}{l}
A=\frac{1}{2} \times \ell r=\frac{1}{2}(20-2 r) \cdot r \\
A=10 r-r^{2}
\end{array}
\)
\(
\begin{array}{l}
\frac{\mathrm{d} \mathrm{A}}{\mathrm{dr}}=10-2 \mathrm{r} \text { and } \frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dr}^{2}}=-2 < 0 \\
\text { When } \frac{\mathrm{d} \mathrm{A}}{\mathrm{dr}}=0 \Rightarrow 10-2 \mathrm{r}=0 \Rightarrow \mathrm{r}=5
\end{array}
\)
Hence area of flower bed will be maximum when \(r=5\).
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