MHT CET · Maths · Application of Derivatives
The maximum area of the rectangle that can be inscribed in a circle of radius \(r\), is
- A \(\pi r^{2}\)
- B \(r^{2}\)
- C \(\pi r^{2} / 4\)
- D \(2 r^{2}\)
Answer & Solution
Correct Answer
(D) \(2 r^{2}\)
Step-by-step Solution
Detailed explanation
Area of rectangle,
\(A =2 x \cdot 2 \sqrt{r^{2}-x^{2}} \)
\( =4 x \sqrt{r^{2}-x^{2}} \)
\( \frac{d A}{d x} =\frac{4\left(r^{2}-2 x^{2}\right)}{\sqrt{r^{2}-x^{2}}} \)
For maximum or minimum put \(\frac{d A}{d x}=0\)
\(
\Rightarrow x=r / \sqrt{2}
\)
It can be easily checked that \(\frac{d^{2} A}{d x^{2}} < 0\) for this value of \(x\). \(\therefore A\) is maximum for \(x=\frac{r}{\sqrt{2}}\) and the maximum value of \(A\) is given by
\(
A=4 \frac{r}{\sqrt{2}} \sqrt{r^{2}-\frac{r^{2}}{2}}=2 r^{2}
\)
\(A =2 x \cdot 2 \sqrt{r^{2}-x^{2}} \)
\( =4 x \sqrt{r^{2}-x^{2}} \)
\( \frac{d A}{d x} =\frac{4\left(r^{2}-2 x^{2}\right)}{\sqrt{r^{2}-x^{2}}} \)
For maximum or minimum put \(\frac{d A}{d x}=0\)
\(
\Rightarrow x=r / \sqrt{2}
\)
It can be easily checked that \(\frac{d^{2} A}{d x^{2}} < 0\) for this value of \(x\). \(\therefore A\) is maximum for \(x=\frac{r}{\sqrt{2}}\) and the maximum value of \(A\) is given by
\(
A=4 \frac{r}{\sqrt{2}} \sqrt{r^{2}-\frac{r^{2}}{2}}=2 r^{2}
\)
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