#### Question

A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area ?

#### Solution

\[\text { Let the dimensions of the rectangle bexandy.Then }, \]

\[\frac{x^2}{4} + y^2 = r^2 \]

\[ \Rightarrow x^2 + 4 y^2 = 4 r^2 \]

\[ \Rightarrow x^2 = 4\left( r^2 - y^2 \right) . . . \left( 1 \right)\]

\[\text { Area of rectangle

}= xy\]

\[ \Rightarrow A = xy\]

\[\text { Squaring both sides, we get }\]

\[ \Rightarrow A^2 = x^2 y^2 \]

\[ \Rightarrow Z = 4 y^2 \left( r^2 - y^2 \right) \left[ \text { From eq } . \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dZ}{dy} = 8y r^2 - 16 y^3 \]

\[\text { For the maximum or minimum values of Z, we must have }\]

\[\frac{dZ}{dy} = 0\]

\[ \Rightarrow 8y r^2 - 16 y^3 = 0\]

\[ \Rightarrow 8 r^2 = 16 y^2 \]

\[ \Rightarrow y^2 = \frac{r^2}{2}\]

\[ \Rightarrow y = \frac{r}{\sqrt{2}}\]

\[\text { Substituting the value ofyineq .} \left( 1 \right), \text { we get }\]

\[ \Rightarrow x^2 = 4\left( r^2 - \left( \frac{r}{\sqrt{2}} \right)^2 \right)\]

\[ \Rightarrow x^2 = 4\left( r^2 - \frac{r^2}{2} \right)\]

\[ \Rightarrow x^2 = 4\left( \frac{r^2}{2} \right)\]

\[ \Rightarrow x^2 = 2 r^2 \]

\[ \Rightarrow x = r\sqrt{2}\]

\[\text{ Now, }\]

\[\frac{d^2 Z}{d y^2} = 8 r^2 - 48 y^2 \]

\[ \Rightarrow \frac{d^2 Z}{d y^2} = 8 r^2 - 48\left( \frac{r^2}{2} \right)\]

\[ \Rightarrow \frac{d^2 Z}{d y^2} = - 16 r^2 < 0\]

\[\text { So, the area is maximum when x =} r\sqrt{2} \text { and }y = \frac{r}{\sqrt{2}} . \]

\[\text { Area } = xy\]

\[ \Rightarrow A = r\sqrt{2} \times \frac{r}{\sqrt{2}}\]

\[ \Rightarrow A = r^2\]