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# Solution for 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.? - CBSE (Commerce) Class 12 - Mathematics

#### 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

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

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

$\Rightarrow x^2 + 4 y^2 = 4 r^2$

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

$\text { Area of rectangle }= xy$

$\Rightarrow A = xy$

$\text { Squaring both sides, we get }$

$\Rightarrow A^2 = x^2 y^2$

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

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

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

$\frac{dZ}{dy} = 0$

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

$\Rightarrow 8 r^2 = 16 y^2$

$\Rightarrow y^2 = \frac{r^2}{2}$

$\Rightarrow y = \frac{r}{\sqrt{2}}$

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

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

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

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

$\Rightarrow x^2 = 2 r^2$

$\Rightarrow x = r\sqrt{2}$

$\text{ Now, }$

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

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

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

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

$\text { Area } = xy$

$\Rightarrow A = r\sqrt{2} \times \frac{r}{\sqrt{2}}$

$\Rightarrow A = r^2$

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Solution 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.? Concept: Graph of Maxima and Minima.
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